Fig. depicts an archery target marked with its five scoring regions from the centre outwards as $Gold, Red, Blue, Black$ and $White.$ The $diameter$ of the region representing Gold score is $21\, cm$ and each of the other bands is $10.5 \,cm$ wide. Find the area of each of the five scoring regions. [use $\pi=\frac{22}{7}$]

(N/A) Radius $(r_1)$ of the gold region (i.e.,$1^{st}$ circle) $= \frac{21}{2} = 10.5\, cm$.
Given that each circle is $10.5\, cm$ wider than the previous circle.
Therefore,radius $(r_2)$ of the $2^{nd}$ circle $= 10.5 + 10.5 = 21\, cm$.
Radius $(r_3)$ of the $3^{rd}$ circle $= 21 + 10.5 = 31.5\, cm$.
Radius $(r_4)$ of the $4^{th}$ circle $= 31.5 + 10.5 = 42\, cm$.
Radius $(r_5)$ of the $5^{th}$ circle $= 42 + 10.5 = 52.5\, cm$.
Area of gold region $= \pi r_1^2 = \frac{22}{7} \times (10.5)^2 = 346.5\, cm^2$.
Area of red region $= \pi r_2^2 - \pi r_1^2 = \frac{22}{7} \times (21^2 - 10.5^2) = 1039.5\, cm^2$.
Area of blue region $= \pi r_3^2 - \pi r_2^2 = \frac{22}{7} \times (31.5^2 - 21^2) = 1732.5\, cm^2$.
Area of black region $= \pi r_4^2 - \pi r_3^2 = \frac{22}{7} \times (42^2 - 31.5^2) = 2425.5\, cm^2$.
Area of white region $= \pi r_5^2 - \pi r_4^2 = \frac{22}{7} \times (52.5^2 - 42^2) = 3118.5\, cm^2$.
Thus,the areas of the $Gold, Red, Blue, Black,$ and $White$ regions are $346.5\, cm^2, 1039.5\, cm^2, 1732.5\, cm^2, 2425.5\, cm^2,$ and $3118.5\, cm^2$ respectively.