9. Circular measure

Learning objectives

9.1Solve arc length and sector area problems• Solve problems involving arc length and sector area of a circle, including knowledge and use of radian measure.• Use radian measure in problems that may involve compound shapes.• Formulas are not given.

An angle of $\theta$ radians is defined by $\theta=s/r$ , where $s$ is the subtended arc length and $r$ is the radius. Because a full circumference is $2\pi r$ , one full turn is $2\pi$ radians and $180^\circ=\pi$ radians.

Quantity, with $\theta$ in radians	Formula
arc length	$s=r\theta$
sector area	$A=\tfrac12r^2\theta$
sector perimeter	$P=2r+r\theta$
triangle made by two radii	$A_\triangle=\tfrac12r^2\sin\theta$
minor segment area	$A_\text{segment}=\tfrac12r^2(\theta-\sin\theta)$

A sector of radius $24$ has area $432$ . From $\tfrac12(24)^2\theta=432$ , $\theta=3/2$ radians. Its arc is therefore $s=24(3/2)=36$ . This also follows from $A=\tfrac12rs$ , which is obtained by combining the first two formulas.

For an annular sector with outer radius $8$ , inner radius $5$ and angle $\pi/3$ , subtract sector areas: $A=\tfrac12(8^2-5^2)(\pi/3)=13\pi/2$ . Its boundary contains both arcs and two radial gaps, so $P=8\pi/3+5\pi/3+2(8-5)=13\pi/3+6$ .

For any compound shape, mark each boundary piece before adding the perimeter, then decompose the area into sectors, triangles, rectangles or segments and attach a plus or minus sign to each. Use $2\pi-\theta$ for a major angle and keep exact $\pi$ or radical values until the final step.

The formulas $s=r\theta$ and $A=\tfrac12r^2\theta$ require radians. A sector perimeter includes two radii; a segment area is sector minus triangle, not sector alone. Convert degrees before substitution and include units: length for arcs and squared units for areas.

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