Figure 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.

Hint:Here, all the five regions are in a circle shape. The area of the gold region is calculated by the area of a circle, which is given by $A = \pi {r^2}$, where r is the radius of the circle. While the formula of area of a ring is used to find out the areas of other four regions i.e., red, blue, black and white, which is given by $\pi \left( {{R^2} - {r^2}} \right)$, where $R$ is the radius of outer ring and $r$ is the radius of inner ring.Using this formula we can calculate the area of each of the five scoring regions.


Complete step-by-step answer:

Given, Diameter of gold region= $21$$cm$
Therefore, Radius of gold region =$\dfrac{{21}}{2}$$cm$ $ = 10.5$$cm$
Area of gold region= $\pi {r^2}$
$ = \pi {\left( {10.5} \right)^2} = \dfrac{{22}}{7} \times 110.25 = 346.5$$c{m^2}$
Hence, Area of gold region= $346.5c{m^2}$
Now, the radius of the red region is calculated by adding the width of the band in the radius of the previous region.
Radius of red region= Radius of gold region+ width of band
$ = 10.5 + 10.5 = 21cm$
Also, the area of red region is calculated by the formula of area of a ring, which is given by $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring .
The radius of a particular region is the radius of outer ring and radius of previous region works as the radius of inner ring in calculating the area of a ring.
Area of red region= \[\pi \left( {{{21}^2} - {{10.5}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {441 - 110.25} \right) = \dfrac{{22}}{7} \times 330.75 = 1039.5c{m^2}$
Hence, Area of red region=$1039.5c{m^2}$

Similarly as the red region, we can calculate the radius and area of the remaining three regions, i.e., blue, black and white.
Radius of blue region= Radius of red region+ width of band
$ = 21 + 10.5 = 31.5cm$
We know that, Area of a ring= $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring
Area of blue region= \[\pi \left( {{{31.5}^2} - {{21}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {992.25 - 441} \right) = \dfrac{{22}}{7} \times 551.25 = 1732.5c{m^2}$
Hence, Area of blue region=$1732.5c{m^2}$

Radius of black region= Radius of blue region+ width of band
= $31.5 + 10.5 = 42cm$
We know that, Area of a ring= $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring
Area of black region= \[\pi \left( {{{42}^2} - {{31.5}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {1764 - 992.25} \right) = \dfrac{{22}}{7} \times 771.75 = 2425.5c{m^2}$
Hence, Area of black region=$2425.5c{m^2}$

Radius of white region= Radius of black region+ width of band
$ = 42 + 10.5 = 52.5cm$
We know that, Area of a ring= $\pi \left( {{R^2} - {r^2}} \right)$, where $R = $radius of outer ring and $r$=radius of inner ring
Area of white region= \[\pi \left( {{{52.5}^2} - {{42}^2}} \right)\]
$ = \dfrac{{22}}{7}\left( {2756.25 - 1764} \right) = \dfrac{{22}}{7} \times 992.25 = 3118.5c{m^2}$
Hence, Area of white region=$3118.5c{m^2}$

Note:While calculating the radius of red, blue, black or white region; it is necessary to add the width of the band in the radius of the previous region. This radius will act as the radius of outer ring and radius of previous region works as the radius of inner ring in calculating the area of a ring.