Question

Two concentric circles have radii of 1.2m and 2.8m. the central angle is 60grad. find the area of the portion of the sector of the larger circle, which is outside the circle. The final answer must be 3.35m²

Answer 1

let's define:

$S = \pi \cdot r^2 \\ \Delta S = S_1 - S_2 = \pi \cdot 2.8^2 - \pi \cdot 1.2^2 = \pi \cdot (2.8^2 - 1.2^2) = \\ \pi \cdot (2.8 - 1.2) (2.8 + 1.2) = \pi \cdot 1.6 \cdot 4 = 6.4 \pi$

It is an area of outside circle without internal circle.

And for central angle of 60 grad it is enough to take 1/6 of this square (as 60 grad sector of a circle is 1/6 of whole circle)

So: $\Delta S_{60 \degree} = \frac{\Delta S}{6} = \frac{6.4 \pi}{6} = \frac{6.4 \cdot 3.1416}{6} = \frac{20.1062}{6} = 3.35 m^2$