Circular segments and annular sectors

For a minor central angle \(0<\theta<\pi\), the minor segment is the sector with triangle \(OAB\) removed. Its area is \(\dfrac12r^2\theta-\dfrac12r^2\sin\theta\), with radians used for both \(\theta\) and the sine calculation. With degrees, use the degree-sector formula and degree-mode sine instead.

Example

A circle has radius 6 cm and minor central angle \(60^\circ\). Minor segment area \(=\dfrac{60}{360}\pi6^2-\dfrac12(6)(6)\sin60^\circ=6\pi-9\sqrt3\text{ cm}^2\). Major segment area is \(36\pi-(6\pi-9\sqrt3)=30\pi+9\sqrt3\text{ cm}^2\).

A segment’s perimeter uses its arc and chord, not two radii. In this example the chord is 6 cm and the minor arc is \(2\pi\text{ cm}\), giving perimeter \(6+2\pi\text{ cm}\).

For an annular sector between concentric radii \(R,r\) with common angle \(\theta\) radians, area is \(\dfrac12(R^2-r^2)\theta\). For \(R=7\text{ cm}\), \(r=4\text{ cm}\), \(\theta=1.2\), area is \(19.8\text{ cm}^2\). The boundary has two arcs and two radial pieces, so perimeter is \(7(1.2)+4(1.2)+2(7-4)=19.2\text{ cm}\).

At a diameter the two segments are semicircles. To find a major segment, subtract the minor segment from the whole circle; do not insert a reflex angle into the minor-segment recipe without analysing the region.