The fig. depicts an archery target marked with its five scoring regions from the centre outwards as Gold,

Radius of the Gold scoring region

$=\frac{21}{2} \mathrm{~cm}=10.5 \mathrm{~cm}(\because$ Diameter $=21 \mathrm{~cm})$

Therefore, the area of the Gold scoring region (circle)

$=\pi \times\left(\frac{21}{2}\right)^{2} \mathrm{~cm}^{2}=\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \mathrm{~cm}^{2}$

$=\frac{\mathbf{3 3} \times \mathbf{2 1}}{\mathbf{2}} \mathbf{c m}^{\mathbf{2}}=\frac{\mathbf{6 9 3}}{\mathbf{2}} \mathbf{c m}^{\mathbf{2}}=346.5 \mathrm{~cm}^{2}$

Radius of the combined circular region scoring Gold and Red

= radius of Gold scoring region

width of the Red scoring band

= 10.5 cm + 10.5 cm = 21 cm

The area of the Red scoring region

= Combined area of Gold and Red scoring region – area of Gold scoring region.

$=\left\{\pi \times(21)^{2}-\pi \times\left(\frac{21}{2}\right)^{2}\right\} \mathrm{cm}^{2}$

$=\pi \times\left\{(21)^{2}-\left(\frac{21}{2}\right)^{2}\right\} \mathrm{cm}^{2}$

$=\frac{22}{7} \times\left\{\left(\frac{21}{2} \times 2\right)^{2}-\left(\frac{21}{2}\right)^{2}\right\} \mathrm{cm}^{2}$

$=\frac{22}{7} \times\left(\frac{21}{2}\right)^{2} \times\{4-1\} \mathrm{cm}^{2}=\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times 3 \mathrm{~cm}^{2}$

$=\frac{\mathbf{6 9 3}}{\mathbf{2}} \times \mathbf{3} \mathbf{c m}^{2}=\frac{\mathbf{2 0 7 9}}{\mathbf{2}} \mathbf{c m}^{2}=1039.5 \mathrm{~cm}^{2}$

Radius of the combined circular region scoring Gold, Red and Blue.

= 10.5 cm + 10.5 cm + 10.5 cm = 31.5 cm

$=\frac{\mathbf{6 3}}{\mathbf{2}} \mathrm{cm}$

Then, area of the Blue scoring region = {Combined area of Gold, Red and Blue scoring regions}

– {Combined area of Gold and Red scoring regions}

$=\pi \times\left(\frac{\mathbf{6 3}}{\mathbf{2}}\right)^{\mathbf{2}}-\pi \times(21)^{2}=\pi \times(21)^{2} \times\left\{\frac{\mathbf{9}}{\mathbf{4}}-\mathbf{1}\right\}$

$=\frac{22}{7} \times 21 \times 21 \times \frac{5}{4} \mathrm{~cm}^{2}=\frac{11 \times 63 \times 5}{2} \mathrm{~cm}^{2}$

$=1732.5 \mathrm{~cm}^{2}$

Similarly, we find the area of the black scoring region

$=\left\{\pi \times(31.5+10.5)^{2}-\pi \times(31.5)^{2}\right\} \mathrm{cm}^{2}$

$=\left\{\pi \times(42)^{2}-\pi \times\left(\frac{63}{2}\right)^{2}\right\} \mathrm{cm}^{2}$

$=\pi \times(21)^{2} \times\left\{4-\frac{9}{4}\right\} \mathrm{cm}^{2}=\frac{22}{7} \times 21 \times 21 \times \frac{7}{4} \mathrm{~cm}^{2}$

$=\frac{231 \times 21}{2} \mathbf{c m}^{2}=\frac{4851}{2} \mathbf{c m}^{2}=2425.5 \mathrm{~cm}^{2}$

Now, the area of the white scoring region

$=\left\{\pi \times(42+10.5)^{2}-\pi \times(42)^{2}\right\} \mathrm{cm}^{2}$

$=\left\{\pi \times\left(\frac{105}{2}\right)^{2}-\pi \times(42)^{2}\right\} \mathrm{cm}^{2}$$=\pi \times(21)^{2} \times\left\{\frac{25}{4}-4\right\} c m^{2}$

$=\frac{22}{7} \times 21 \times 21 \times \frac{9}{4} \mathrm{~cm}^{2}$

$=\frac{6237}{2} \mathbf{m m}^{2}=3118.5 \mathrm{~cm}^{2}$