Mathematics/Lessons/Trigonometry/Areas of Sectors and Segments
Pure · Trigonometry · Year 2

Areas of Sectors and Segments

The same proportion argument that gave a clean formula for arc length works just as well for area — and one extra step, subtracting a triangle, unlocks the area cut off by a chord.

8 min read 3 worked examples

Sector area

The area of a sector (a "pizza slice" of the circle) with angle $\theta$ measured in radians is:

$A = \dfrac{1}{2}r^2\theta$
DerivationSector area as a proportion of the full circle
1
The full area of a circle of radius $r$ is $\pi r^2$, corresponding to the full angle $2\pi$ radians.
Standard area formula, over the full turn.
2
A sector of angle $\theta$ is the same proportion of the full area: $\dfrac{A}{\pi r^2} = \dfrac{\theta}{2\pi}$.
Sector area scales directly with the angle it subtends, exactly as arc length did in the previous lesson.
3
Multiply both sides by $\pi r^2$: $A = \pi r^2 \cdot \dfrac{\theta}{2\pi} = \dfrac{1}{2}r^2\theta$.
The $\pi$ cancels, leaving the compact formula.
▪ QED

Worked example 1

Sector area, finding the radius
A sector of a circle has angle $1.2$ radians and area $54\text{ cm}^2$. Find the radius of the circle.
MethodSubstitute into $A=\frac12 r^2\theta$ and solve for $r$
1
$54 = \dfrac{1}{2}r^2(1.2)$.
Substitute the given area and angle into the sector area formula.
2
$54 = 0.6r^2 \Rightarrow r^2 = \dfrac{54}{0.6} = 90$.
Simplify and isolate r².
3
$r = \sqrt{90} = 3\sqrt{10} \approx 9.49\text{ cm}$ (3 s.f.).
Take the positive square root, since radius must be positive.
▪ solved
Examiner's eye

Keep the angle in radians throughout — plugging $\theta=1.2$ straight into the formula only works because it's already in the correct unit. If a question instead gives the angle in degrees, convert to radians as the very first step.

Segment area

A chord — a straight line joining two points on a circle — divides the circle into two regions. The smaller region, bounded by the chord and the minor arc, is called a segment. A segment is not a sector: it's what's left of a sector once the triangle formed by the two radii and the chord is cut away.

$\text{segment area} = \text{sector area} - \text{triangle area}$
A circle centred at O with two radii OA and OB meeting at angle theta, and the chord AB drawn between them. The sector OAB is split into two regions by the chord: the triangle OAB below the chord, shaded gold, and the segment above the chord between the chord and the minor arc AB, shaded teal.
Sector $OAB$ splits along the chord $AB$ into two pieces: triangle $OAB$ (gold) and the segment (teal). Subtract one from the whole sector to get the other.
DerivationSegment area via the sector-minus-triangle idea
1
The sector $OAB$ (angle $\theta$, radius $r$) has area $\dfrac12r^2\theta$.
The sector area formula, exactly as above.
2
Triangle $OAB$ has two sides of length $r$ with included angle $\theta$, so its area is $\dfrac12r^2\sin\theta$.
The Year 1 "$\frac12ab\sin C$" area formula, with $a=b=r$ and $C=\theta$.
3
Segment area $= \dfrac12r^2\theta - \dfrac12r^2\sin\theta = \dfrac12r^2(\theta-\sin\theta)$.
Subtract the triangle from the sector and factor out $\frac12r^2$.
▪ QED
Why this needs radians even more than arc length does

$A_{\text{segment}}=\dfrac12r^2(\theta-\sin\theta)$ mixes $\theta$ used two different ways in the same expression — once directly (from the sector) and once inside $\sin\theta$ (from the triangle). Both uses require $\theta$ in radians; there's no way to patch this formula up for a degree-valued angle.

Worked example 2

Segment area
A sector $OAB$ has radius $9\text{ cm}$ and angle $\dfrac{2\pi}{3}$ radians. Find the area of the segment cut off by the chord $AB$, correct to 3 significant figures.
A circle centred at O with radius 9. Two radii OA and OB meet at angle 2 pi over 3, with the chord AB between A and B. The segment above the chord, between the chord and the minor arc AB, is shaded teal.
The shaded segment is the region to find — bounded by chord $AB$ and the minor arc, not the whole sector.
MethodApply $A=\frac12r^2(\theta-\sin\theta)$ directly
1
$r=9$, $\theta=\dfrac{2\pi}{3}$, and $\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$ (an exact Year 1 value).
Identify what's needed before substituting — the exact sine value keeps the working clean.
2
$A = \dfrac12(9)^2\left(\dfrac{2\pi}{3}-\dfrac{\sqrt3}{2}\right) = \dfrac{81}{2}\left(\dfrac{2\pi}{3}-\dfrac{\sqrt3}{2}\right)$.
Substitute directly into the segment area formula.
3
$A = 27\pi - \dfrac{81\sqrt3}{4} \approx 49.7\text{ cm}^2$ (3 s.f.).
Expand and simplify; the exact form is a valid answer in its own right.
▪ solved

Worked example 3

Perimeter of a segment · Challenge
For the same sector as Example 2 ($r=9\text{ cm}$, $\theta=\dfrac{2\pi}{3}$), find the exact perimeter of the segment cut off by the chord $AB$.
The same circle as before, radius 9, angle 2 pi over 3 at O. The arc AB and the chord AB are both drawn solid and black, labelled s equals 6 pi and AB equals 9 root 3. The two radii OA and OB are drawn dashed and grey to show they are not part of the segment's boundary.
The segment's perimeter is the arc (orange) plus the chord (teal) only — the two radii (dashed) bound the triangle, not the segment.
MethodAdd the arc length to the chord length, not to the two radii
1
A segment's boundary is the arc $AB$ plus the straight chord $AB$ — unlike a sector, it has no radii on its boundary.
The two radii $OA$, $OB$ are internal to the sector, not part of the segment's edge.
2
Arc length: $s=r\theta = 9\times\dfrac{2\pi}{3} = 6\pi$.
The standard arc length formula.
3
Chord length: splitting triangle $OAB$ down the middle gives $AB = 2r\sin\left(\dfrac{\theta}{2}\right) = 2(9)\sin\left(\dfrac{\pi}{3}\right) = 18\times\dfrac{\sqrt3}{2} = 9\sqrt3$.
Bisecting the isosceles triangle formed by the two radii and the chord gives two right-angled triangles, each with hypotenuse $r$ and angle $\theta/2$.
4
Perimeter $= 6\pi + 9\sqrt3$ cm (exact).
Add the curved arc to the straight chord.
▪ solved
Chord length, on demand

$AB=2r\sin\left(\dfrac{\theta}{2}\right)$ is worth remembering as a standalone result — it turns up any time a question needs the straight-line distance between two points on a circle, not just when finding a segment's perimeter.

Common mistake

Treating a segment's perimeter as arc $+$ two radii (copying the sector formula) — a segment has no radii on its boundary at all. Equally common: computing "segment area" as $\frac12r^2\theta$ alone, forgetting to subtract the triangle. Always ask whether the shape includes the two straight radii (a sector) or is bounded only by the chord (a segment).

Check yourself

A sector has area $40\text{ cm}^2$ and angle $0.8$ radians. Find the radius, and hence find the area of the segment cut off by the corresponding chord, correct to 3 significant figures.
(Hint: find r from $A=\frac12r^2\theta$ first, then apply $A_{\text{segment}}=\frac12r^2(\theta-\sin\theta)$.)

A circle centred at O with radius r (unknown). Two radii OA and OB meet at angle 0.8 radians. The chord AB is drawn, and the segment between the chord and the minor arc AB is shaded teal.
The unknown radius $r$ and the given angle $\theta=0.8$ — the shaded segment is what the second part asks for.
Method
1
$40 = \dfrac12r^2(0.8) = 0.4r^2 \Rightarrow r^2 = 100 \Rightarrow r=10\text{ cm}$.
Rearrange the sector area formula for r, then take the positive square root.
2
$A_{\text{segment}} = \dfrac12(10)^2(0.8-\sin0.8) = 50(0.8-\sin0.8)$.
Substitute r and θ into the segment area formula.
3
$\sin0.8 \approx 0.7174$, so $A_{\text{segment}} \approx 50(0.0826) \approx 4.13\text{ cm}^2$ (3 s.f.).
Evaluate with a calculator in radian mode — this is the extra check worth making every time.
▪ solved
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