What Is a Segment of a Circle?
A segment of a circle is the region bounded by a chord and the arc that the chord cuts off. A chord is any straight line joining two points on the circle; the arc is the curved part of the circle's edge between those same two points. Together they enclose the segment.
Every chord (unless it passes through the centre) splits the disc into two segments:
The minor segment — the smaller region, cut off by the minor arc (the shorter arc).
The major segment — the larger region, bounded by the major arc (the longer arc).
It helps to keep a segment distinct from its close relative, the sector. A sector is the "pizza slice" bounded by two radii and an arc — it reaches the centre. A segment is bounded by a chord and an arc — it does not reach the centre. The difference between a sector and the triangle inside it is the segment, and that single relationship generates the whole area formula.
Is a segment the same as a sector?
No, and the confusion is the most-asked question on the topic. A sector includes the centre and is enclosed by two radii; a segment is cut off by a single straight chord and leaves the centre out. Picture the sector as the full pizza slice and the segment as just the rounded tip beyond the straight chord. This article uses the parts of a circle — chord, arc, radius — as its building blocks throughout.
The Area of a Segment of a Circle
Here is where the sector-minus-triangle idea pays off. To find the area of the minor segment, take the sector that contains it and subtract the triangle formed by the two radii and the chord. What remains is exactly the segment.
Area of segment = Area of sector − Area of triangle
Each piece has a known formula. The sector area for a central angle $\theta$ in degrees is $\frac{\theta}{360°} \times \pi r^2$, since the sector is that fraction of the full circle. The triangle formed by the two radii (each of length $r$) with included angle $\theta$ has area $\frac{1}{2} r^2 \sin\theta$. Subtracting gives the segment:
$$\text{Area of minor segment} = \frac{\theta}{360°} \pi r^2 - \frac{1}{2} r^2 \sin\theta$$
When the angle $\theta$ is measured in radians, the sector fraction simplifies to $\frac{1}{2}r^2\theta$, so the formula becomes cleaner:
$$\text{Area of minor segment} = \frac{1}{2} r^2 (\theta - \sin\theta)$$
Variable key: $r$ is the radius of the circle; $\theta$ is the central angle subtended by the chord (the angle ∠AOB at the centre); the answer is in square units. The $\sin\theta$ term needs $\theta$ in the matching mode — degrees for the first formula, radians for the second.
Why does the radian formula drop the $360$?
Because radians measure angle as a fraction of $2\pi$ rather than $360°$, the sector fraction $\frac{\theta}{2\pi}\times\pi r^2$ simplifies to $\frac{1}{2}r^2\theta$ — the $\pi$ and the $360$ both disappear. That is the whole reason the radian version looks cleaner; it is the same formula, written in the angle unit that cancels.
The major segment
The major segment is everything left over, so its area is the whole circle minus the minor segment:
$$\text{Area of major segment} = \pi r^2 - \left(\frac{\theta}{360°} \pi r^2 - \frac{1}{2} r^2 \sin\theta\right)$$
Perimeter of a segment
The boundary of a segment is the arc plus the chord. Using the arc-length formula and the chord length:
$$\text{Perimeter} = \underbrace{r\theta}{\text{arc (radians)}} + \underbrace{2r\sin\frac{\theta}{2}}{\text{chord}}$$
The arc length here uses the standard arc length relation; the chord length comes from the chords of a circle.
Examples of Segment of a Circle
The examples build from a clean radian case to a degree problem and a major-segment question. Each states its units.
Example 1
A circle has radius 10 cm. A chord subtends a central angle of 2 radians. Find the area of the minor segment. Take $\sin 2 \approx 0.909$.
$$\text{Area} = \frac{1}{2} r^2 (\theta - \sin\theta) = \frac{1}{2} \times 10^2 \times (2 - 0.909)$$
$$\text{Area} = 50 \times 1.091 = 54.55 \text{ cm}^2$$
The minor segment is about 54.55 square centimetres.
Example 2
A circle of radius 6 cm has a chord cutting off a 60° segment. Find the minor segment area — and watch the common slip first. Use $\pi \approx 3.14$ and $\sin 60° \approx 0.866$.
The instinct is often to compute the sector and call that the segment. The sector area is $\frac{60}{360} \times 3.14 \times 36 = 18.84$ cm². But the sector includes the triangle, and the segment does not — so 18.84 must be too big. The check: the segment can only be smaller than its sector, never equal to it.
Subtracting the triangle fixes it:
$$\text{Triangle} = \frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times 36 \times 0.866 = 15.59 \text{ cm}^2$$
$$\text{Segment} = 18.84 - 15.59 = 3.25 \text{ cm}^2$$
The minor segment is about 3.25 square centimetres — far smaller than the sector, as it must be.
Example 3
A circle of radius 14 cm has a chord subtending 90° at the centre. Find the minor segment area. Use $\pi \approx \frac{22}{7}$.
$$\text{Sector} = \frac{90}{360} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times \frac{22}{7} \times 196 = 154 \text{ cm}^2$$
$$\text{Triangle} = \frac{1}{2} \times 14^2 \times \sin 90° = \frac{1}{2} \times 196 \times 1 = 98 \text{ cm}^2$$
$$\text{Segment} = 154 - 98 = 56 \text{ cm}^2$$
Example 4
For the same circle in Example 3 (radius 14 cm, 90° chord), find the major segment area.
The whole circle has area $\frac{22}{7} \times 196 = 616$ cm². The major segment is the circle minus the minor segment:
$$\text{Major segment} = 616 - 56 = 560 \text{ cm}^2$$
Example 5
Find the perimeter of the minor segment in Example 1 (radius 10 cm, central angle 2 radians). Take $\sin 1 \approx 0.841$.
$$\text{Arc} = r\theta = 10 \times 2 = 20 \text{ cm}$$
$$\text{Chord} = 2r\sin\frac{\theta}{2} = 2 \times 10 \times \sin 1 = 20 \times 0.841 = 16.82 \text{ cm}$$
$$\text{Perimeter} = 20 + 16.82 = 36.82 \text{ cm}$$
Example 6
A horizontal cylindrical tank of radius 1 m is filled so the water surface is a chord subtending 120° at the centre, measured from the top. The water sits in the major segment below. Find the cross-sectional area of the water. Use $\pi \approx 3.14$ and $\sin 120° \approx 0.866$.
The empty region above the water is the minor segment for $\theta = 120°$:
$$\text{Minor segment} = \frac{120}{360} \times 3.14 \times 1^2 - \frac{1}{2} \times 1^2 \times 0.866 = 1.047 - 0.433 = 0.614 \text{ m}^2$$
The water fills the rest of the disc:
$$\text{Water area} = \pi r^2 - 0.614 = 3.14 - 0.614 = 2.53 \text{ m}^2$$
The water's cross-section is about 2.53 square metres — the kind of tank problem that opened this article.
Why Segments Show Up Everywhere Round
A segment is what a straight line leaves behind in a round region, and round regions cut by straight lines are everywhere.
Tank gauging — the area of liquid in a horizontal cylindrical tank is a circular segment; reading a dipstick into a volume relies on exactly this formula.
Architecture and design — the glass in an arched ("segmental") window, the area of a curved balcony, or the steel in a curved bridge plate all reduce to segment areas.
Engineering cross-sections — pipes, channels, and culverts running part-full are sized using the segment of the circle their walls trace.
The underlying reason the segment matters is that the area of a circle on its own answers only "how much in total" — the segment answers "how much on one side of a straight cut", which is the question real cross-sections ask. The relationship between a chord, its arc, and the angle it subtends was already understood by Archimedes, whose work on areas bounded by curves laid the groundwork for integral calculus nearly two thousand years later.
Where Segment Problems Trip Students Up
Mistake 1: Confusing a segment with a sector
Where it slips in: The very first step, when a student reads "segment" and pictures the pizza-slice sector instead.
Don't do this: Report the sector area $\frac{\theta}{360°}\pi r^2$ as the segment area.
The correct way: A segment leaves the centre out — it is the sector minus the triangle made by the two radii and the chord. The first-instinct error is treating the two as the same shape; the giveaway is that a segment must always come out smaller than its sector.
Mistake 2: Mixing degree and radian formulas
Where it slips in: Choosing the area formula when the angle is given in radians but the degree formula is the one memorised.
Don't do this: Plug a radian angle into $\frac{\theta}{360°}\pi r^2 - \frac{1}{2}r^2\sin\theta$, or set the calculator to the wrong mode for the $\sin\theta$ term.
The correct way: Match the formula to the angle. Radians use $\frac{1}{2}r^2(\theta - \sin\theta)$ with the calculator in radian mode; degrees use the $\frac{\theta}{360°}$ version in degree mode. The memorizer who recalls one formula but not which angle mode it needs is exactly who lands here — the $\sin\theta$ term is where the mode error hides.
Mistake 3: Forgetting the chord when finding the perimeter
Where it slips in: Perimeter questions, where students compute the arc and stop.
Don't do this: Report the arc length $r\theta$ alone as the segment's perimeter.
The correct way: A segment's boundary is the arc plus the chord, so add $2r\sin\frac{\theta}{2}$. The rusher who treats "perimeter" as "arc length" drops the straight side entirely.
Conclusion
A segment of a circle is the region between a chord and the arc it cuts off, leaving the centre out (unlike a sector).
A chord splits the disc into a smaller minor segment and a larger major segment.
The area of a minor segment is the sector minus the triangle: $\frac{\theta}{360°}\pi r^2 - \frac{1}{2}r^2\sin\theta$ in degrees, or $\frac{1}{2}r^2(\theta - \sin\theta)$ in radians.
The major segment is the whole circle minus the minor segment.
The most common mistakes are confusing a segment with a sector and mixing degree and radian formulas.
Practice and Next Steps
Work through these problems to solidify your understanding, then check each against the formulas above.
A circle of radius 8 cm has a chord subtending 90° at the centre. Find the minor segment area ($\pi \approx 3.14$).
Find the major segment area for the same circle and chord.
A circle of radius 5 cm has a chord subtending $\frac{\pi}{2}$ radians. Find the perimeter of the minor segment.
To work through circle geometry with a teacher who shows where each formula comes from, explore Bhanzu's geometry tutor, our middle school math tutor, or math classes online. Want a live Bhanzu trainer to walk through more segment problems? Book a free demo class.
Read More
Circumference of a circle — the distance around a circle that a segment's arc is part of.
Semicircle — the special segment formed when the chord is a diameter.
Secant of a circle — the line that extends a chord beyond the circle.
Center of a circle — the point the central angle is measured from.
Radius of a circle — the centre-to-edge length every circle formula relies on.
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