Segments and annular sectors

Circular segments

For \(0<\theta<\pi\), the minor segment is the minor sector with triangle \(OAB\) removed.

\(A_{\text{minor segment}}=\dfrac12r^2\theta-\dfrac12r^2\sin\theta=\dfrac12r^2(\theta-\sin\theta)\).

Use radians for both \(\theta\) and \(\sin\theta\) in this formula. Find a major segment by subtracting the minor segment from \(\pi r^2\).

Annular sectors

The shaded region lies between two concentric arcs. The two sectors have the same angle \(\theta\) radians.

\(A_{\text{annular sector}}=\dfrac12R^2\theta-\dfrac12r^2\theta=\dfrac12(R^2-r^2)\theta\).

Its boundary contains two arcs and two straight pieces: \(P=R\theta+r\theta+2(R-r)\).

Review check

I can identify the actual boundary, choose the angle unit, distinguish a sector from a segment, and explain which area is added or removed. I can work backwards from a length or area and check the units and accuracy of my answer.