Fig. 12.3 depicts an archery target marked with its five scoring areas 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.

Explanation

Radius `(r_1)` of the gold region = `21/2` cm = `10.5` cm
Given that each circle is `10.5` cm wider than the previous circle.
Therefore, radius `(r_2)` of the `2nd` circle = `10.5` cm + `10.5` cm = `21` cm
Radius `(r_3)` of the `3rd` circle = `21` cm + `10.5` cm = `31.5 `cm
Radius `(r_4)` of the `4 th` circle = `31.5 cm + 10.5 cm = 42` cm
Radius `(r_5)` of the `5 th` circle = `42 cm + 10.5 cm = 52.5 `cm
Area of the gold region = Area of the 1st circle = `pir₁^2 = pi (10.5)^2 = 346.5 cm^2`
Area of the red region = Area of the `2nd` circle - Area of the `1st` circle
= `pir_2^2 - pir_1^2`
= `π(21)^2 - π(10.5)^2 cm^2`

= `(1386 - 346.5) cm^2`

= `1039.5 cm^2`
Area of blue region = Area of 3rdcircle - Area of 2ndcircle
= `pir_3^2 - pir_2^2`
= `pi(31.5)^2 -pi(21)^2 cm^2`
= `992.25pi - 441pi cm^2 `
= `551.25pi cm^2 `
= `1732.5 cm^2 `

Area of black region = Area of 4thcircle - Area of 3rd circle.
= `pir_4^2 - pr_3^2`
= `pi(42)^2 - pi(31.5)^2`
= `1764pi - 992.25pi`
= `771.75pi`
= `2425.5 cm²`
Area of white region = Area of 5thcircle - Area of 4 thcircle
=` pir_5^2 - pir_4^2`
= `pi(52.5)^2 - pi(42)^2`
= `2756.25pi - 1764pi`
= `992.25pi`
= `3118.5 cm²`
Hence, 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.