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Sector area is the fraction of the circle’s total area corresponding to the central angle:

Asector=θ360∘⋅πr2.A_{\text{sector}}=\frac{\theta}{360^\circ}\cdot \pi r^2.

Here r=9r=9 and θ=80∘\theta=80^\circ:

Asector=80360⋅π⋅92=29⋅81π=18π cm2.A_{\text{sector}}=\frac{80}{360}\cdot \pi\cdot 9^2 = \frac{2}{9}\cdot 81\pi = 18\pi\ \text{cm}^2.

This problem targets CLI’s Geometry objectives on circles—moving comfortably among radius, angle measure, and area. Note the distinction between sector area (uses πr2\pi r^2) and arc length (uses 2πr2\pi r); students sometimes mix the two. Proportional reasoning is key: 80∘80^\circ is 29\tfrac{2}{9} of a full turn, so the sector must be 29\tfrac{2}{9} of the disk’s area. Dimensional analysis supports correctness: areas are in square units; if you accidentally used the circumference formula, you’d get linear units. For mental checks, approximate: π≈3.14⇒18π≈56.5 cm2\pi\approx3.14\Rightarrow 18\pi\approx56.5\ \text{cm}^2, reasonable for a fairly large radius. Mastery of this relationship scales to related tasks such as area of annuli, arc length/sector pairs, and converting degrees to radians—skills frequently assessed in placement to gauge readiness for trigonometry and calculus.