Question

Find the area of a sector with a central angle of 32 degrees and a radius of 8.5 millimeters. Round to the nearest tenth.

Answer 1

$\dfrac{\text{area of circle}}{\text{one full revolution}}=\dfrac{\pi \times8.5^2}{2\pi\text{ rad}}=\dfrac{A}{32^\circ\times\frac{\pi\text{ rad}}{180^\circ}}=\dfrac{\text{area of sector}}{\text{central angle}}$

From this relation it follows that

$A=\dfrac{8.5^2}2\times32\times\dfrac\pi{180}\approx20.2\text{ mm}^2$