A sector of angle theta radians is the fraction theta/(2pi) of a circle: A= 12r2theta. Its arc is s=rtheta, so A= 12rs. Choose the correct boundary first: P=2r+s=r(2+theta). If a minor angle is given but the major sector is used, replace it by 2pi-theta. For inverse problems, write the area and boundary equations, eliminate s or theta, solve for a positive radius, and check 0<theta<2pi for a proper sector. For example, A=18 cm2 and s=6 cm give r=2A/s=6 cm, theta=s/r=1 radian and P=18 cm. Exam check: an arc is only the curved edge. An algebraic positive radius can still imply an angle exceeding one turn; reject it for a proper sector.

A sector of angle \(\theta\) radians is the fraction \(\frac{\theta}{2\pi}\) of a circle: \(A=\frac12r^2\theta\). Its arc is \(s=r\theta\), so \(A=\frac12rs\).

Choose the correct boundary first: \(P=2r+s=r(2+\theta)\). If a minor angle is given but the major sector is used, replace it by \(2\pi-\theta\).

For inverse problems, write the area and boundary equations, eliminate \(s\) or \(\theta\), solve for a positive radius, and check \(0<\theta<2\pi\) for a proper sector.

For example, \(A=18\text{ cm}^2\) and \(s=6\text{ cm}\) give \(r=\frac{2A}{s}=6\text{ cm}\), \(\theta=\frac{s}{r}=1\) radian and \(P=18\text{ cm}\).

Exam check: an arc is only the curved edge. An algebraic positive radius can still imply an angle exceeding one turn; reject it for a proper sector.

Sector area and inverse dimensions

Sector area and inverse dimensions