Figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue , Black and White. The diameter of the region representing Gold score is 21 cm and reach of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

The area of each of the five scoring regions are as under :

For Gold : \(\pi (10.5)^2 \)
\( = \ \frac{22}{7} × 110.25 \)

\( = \ \frac{2425.5}{7} \ \)
\( = \ 346.5 \ cm^2\)

For Red : \( \pi [ (21)^2 \ - \ (10.5)^2 ] \ \)
\( = \ \frac{22}{7}(441 \ - \ 110.25) \ \)
\( = \ \frac{22}{7} × 330.75 \)

\(= \ \frac{7276.5}{7} \ \)
\( = \ 1039.5 \ cm^2 \)

For Blue : \( \pi [(31.5)^2 \ - \ (21)^2] \ \)
\( = \ \frac{22}{7}(992.25 \ - \ 441) \)

\( = \ \frac{22}{7} × 551.25 \ \)
\( = \ \frac{12127.5}{7} \ \)
\( = \ 1732.5 \ cm^2 \)

For Black : \( \pi [(42)^2 \ - \ (31.5)^2] \ \)
\( = \ \frac{22}{7} (1764 \ - \ 992.25) \)

\( = \ \frac{22}{7} × 771.75 \ \)
\( = \ \frac{16978.5}{7} \ \)
\( = \ 2425.5 \ cm^2 \)

For White : \( \pi [(52.5)^2 \ - \ (42)^2] \ \)
\( = \ \frac{22}{7} (2756.25 \ - \ 1764) \)

\(= \ \frac{22}{7} × 992.25 \ \)
\( = \ \frac{21829.5}{7} \ \)
\( = \ 3118.5 \ cm^2 \)