Fig. depicts an archery target marked with its five scoring regions from the center 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.

Let us given that, 

Radius of the gold circle =  r₁ = 21/2 cm = 10.5

Radius of the 2^nd circle = r_{2 =} 10.5 $+$ 10.5  cm = 21 cm

Radius of the 3 ^rd circle =r₃  = 21 $+$ 10.5  cm = 31.5 cm

Radius of the 4^th circle = r₄ = 31.5 $+$ 10.5  cm = 42 cm

Radius of the 5^th circle =r₅  = 42 $+$ 10.5  cm = 52.5 cm

We know that, Area of a circle = $\mathrm{\pi }$r²

 Area of gold circle =  Area of 1^st circle =  $\mathrm{\pi }$r₁²

= $\mathrm{\pi }$ (10.5)² 

= 346.5 cm²

Area of the red region = Area of the 2^nd circle - Area of the 1^st circle

= $\mathrm{\pi }$r₂² - πr₁²

= $\mathrm{\pi }$(21)² - $\mathrm{\pi }$(10.5)² 

= (1386 - 346.5) 

= 1039.5 cm² 

Area of blue region = Area of 3^rdcircle - Area of 2^ndcircle

= $\mathrm{\pi }$r₃² - $\mathrm{\pi }$r₂²

= $\mathrm{\pi }$(31.5)² -$\mathrm{\pi }$(21)² 

= 992.25π - 441$\mathrm{\pi }$

= 551.25$\mathrm{\pi }$

= 1732.5 cm² 

Area of black region = Area of 4^thcircle - Area of 3^rd circle

= $\mathrm{\pi }$r₄ ² - $\mathrm{\pi }$r₃²

= $\mathrm{\pi }$(42)² - $\mathrm{\pi }$(31.5)²

= 1764$\mathrm{\pi }$ - 992.25$\mathrm{\pi }$

= 771.75$\mathrm{\pi }$

= 2425.5 cm²

Area of white region = Area of 5^thcircle - Area of 4 ^thcircle

= $\mathrm{\pi }$r₅ ² - $\mathrm{\pi }$r₄ ²

= $\mathrm{\pi }$(52.5)² - $\mathrm{\pi }$(42)²

= 2756.25$\mathrm{\pi }$ - 1764$\mathrm{\pi }$

= 992.25$\mathrm{\pi }$

= 3118.5 cm²

Therefore, areas of the gold, red, blue, black, and white regions are 346.5 cm², 1039.5 cm², 1732.5 cm², 2425.5 cm² and 3118.5 cm² respectively.