Question

Given 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 each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions. [use π = 22/7]

 

Answer 1

Radius (r₁) of gold region (i.e., 1^st circle)  = 21/2 = 10.5 cm

Given that each circle is 10.5 cm wider than the previous circle.

Therefore, radius (r₂) of 2^nd circle = 10.5 + 10.5

21 cm

Radius (r₃) of 3^rd circle = 21 + 10.5

= 31.5 cm

Radius (r₄) of 4^th circle = 31.5 + 10.5

= 42 cm

Radius (r₅) of 5^th circle = 42 + 10.5

= 52.5 cm

Area of gold region = Area of 1^st circle = `pir_1^2 = pi(10.5)^2 = 346.5 cm^2`

Area of red region = Area of 2^nd circle − Area of 1^st circle

`=pir_2^2-pir_1^2`

`=pi(21)^2 - pi(10.5)^2`

= 441Π - 110.25Π = 330.75Π

= 1039.5 cm²

Area of blue region = Area of 3^rd circle − Area of 2^nd circle

`= pir_3^2 - pi_1^2`

`=pi(31.5)^2 -pi(21)^2`

`=992.25pi - 441pi = 551.25pi`

= 1732.5 cm²

Area of black region = Area of 4^th circle − Area of 3^rd circle

`= pir_4^2 - pir_3^2`

`=pi(42)^2-pi(31.5)^2`

`= 1764pi - 992.25pi`

Area of white region = Area of 5^th circle − Area of 4^th circle

`=pir_5^2 - pir_4^2`

`=pi(52.5)^2-pi(42)^2`

`= 2756.25pi - 1764pi`

`= 992.25pi = 3118.5 cm^2`

Therefore, areas of 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.