Segment Area and Composite Shapes (AA SL)
A segment is the sliver of a circle cut off by a chord, and its area is found by subtracting a triangle from a sector - a formula that isn't in the booklet, so it needs to be built rather than recalled. Composite problems push this further, stitching several sectors (and sometimes shapes outside the circle entirely) into a single perimeter or area. This page covers the segment formula, how to spot when a problem is asking for it, and a fully worked composite example. It's part of the broader Radians, Arcs & Sectors topic.
13 questions on this sub-topic.
The key formulas
Covered under IB syllabus reference SL3.4: the circle - radian measure of angles, length of an arc, area of a sector. Sector area is in the formula booklet; segment area is not - you build it yourself from two booklet pieces.
Sector area
\[A = \tfrac12 r^2\theta\]
Half the radius squared, times the angle in radians.
Segment area
\[\text{Segment area} = \tfrac12r^2\theta - \tfrac12r^2\sin\theta\]
Sector area minus the triangle area, both using the same \(r\) and \(\theta\). Not in the booklet - derived by subtraction.
Need the arc length formula too, or a reminder of the conversion step? See Arc Length and Sector Area and Radians and Conversion.
Worked examples
A circle has radius 4. A sector has central angle \(\tfrac{\pi}{2}\).
Find the exact area of the minor segment.
Worked solution
Sector area: \(A_{\text{sec}}=\tfrac12 r^2\theta=\tfrac12(16)\tfrac{\pi}{2}\) M1 \(=4\pi.\) A1
Triangle area: \(\tfrac12 r^2\sin\theta=\tfrac12(16)\sin\tfrac{\pi}{2}\) M1 \(=8.\) A1
Segment = sector − triangle \(=4\pi-8.\) A1
A goat is tied by a 6 m rope to a corner of a square shed of side 4 m, in an open field. As it grazes around the outside, the rope wraps the corners.
Find the total grazing area (assume it can sweep \(\dfrac{3\pi}{2}\) at the tied corner and \(\dfrac{\pi}{2}\) each at the two adjacent corners with remaining rope 2 m).
Worked solution
Main region: at the tied corner the goat sweeps \(\tfrac{3\pi}{2}\) with the full 6 m rope: \(\tfrac12(6^2)\cdot\tfrac{3\pi}{2}\) M1 \(=27\pi\). A1
Wrapped corners: past each adjacent corner the rope is \(6-4=2\) m and sweeps \(\tfrac{\pi}{2}\); two of them give \(2\times\tfrac12(2^2)\cdot\tfrac{\pi}{2}=2\pi\). M1
Total: \(27\pi+2\pi=29\pi\approx 91.1\) m\(^2\). A1
Common mistakes
- Treating a segment as if it were a sector. The segment area needs the triangle area subtracted from the sector area - just using \(\tfrac12r^2\theta\) alone gives the sector, not the segment.
- Evaluating \(\sin\theta\) in the wrong angle mode. In the segment formula \(\theta\) is already in radians, so \(\sin\theta\) must be evaluated in radian mode on the GDC - switching to degree mode here silently gives a completely different (and much smaller) triangle area.
- Losing track of which pieces build a composite region. In problems with several sectors of different radii (like a rope wrapping around corners), it's easy to add one sector twice or forget that the remaining rope length shortens after each wrap - sketch the regions before combining areas.
Ready to practise properly?
13 segment-area and composite-shape questions, marked instantly like the real exam.
Quick answers
What is the formula for the area of a segment?
\(\text{Segment area} = \tfrac12r^2\theta - \tfrac12r^2\sin\theta\), the sector area minus the area of the triangle formed by the two radii and the chord. It is not in the formula booklet; you derive it by subtraction.
How do I find the perimeter of a sector or segment?
A sector's perimeter is the arc length plus the two radii: \(r\theta + 2r\). A segment's perimeter is just the arc length plus the chord length, since a segment has no straight radii on its boundary.