\(\dfrac{\text{area of circle}}{\text{one full revolution}}=\dfrac{\pi \times8.5^2}{2\pi\text{ rad}}=\dfrac{A}{32^\circ\times\frac{\pi\text{ rad}}{180^\circ}}=\dfrac{\text{area of sector}}{\text{central angle}}\)
From this relation it follows that
\(A=\dfrac{8.5^2}2\times32\times\dfrac\pi{180}\approx20.2\text{ mm}^2\)
