Depucts an archery target market with …

According to the question, 
{tex}r ={/tex} Radius of the region representing Gold score {tex}= 10.5 \ cm{/tex}
r₁ = Radius of the region representing Gold and Red scoring areas
{tex}= (10.5 + 10.5) cm = 21 cm ={/tex}{tex}2r \ cm{/tex}
r₂ = Radius of the region representing Gold, Red and Blue scoring areas
{tex}= (21 + 10.5) cm = 31.5 cm ={/tex}{tex}3r \ cm{/tex}
r₃ = Radius of the region representing Gold, Red, Blue and Black scoring areas
{tex}= (31.5 + 10.5) cm = 42 cm ={/tex} {tex}4r \ cm{/tex}
r₄ = Radius of the region representing Gold, Red, Blue, Black and white scoring areas
{tex}= (42 + 10.5) cm = 52.5 cm ={/tex} {tex}5r \ cm{/tex}
A₁ = Area of the region representing Gold scoring area = {tex} \pi r ^ { 2 } = \frac { 22 } { 7 } \times ( 10.5 ) ^ { 2 } = \frac { 22 } { 7 } \times 10.5 \times 10.5{/tex}
{tex} = 22 \times 1.5 \times 10.5 = 346.5 \mathrm { cm } ^ { 2 }{/tex}
A₂ = Area of the region representing Red scoring area = {tex} \pi ( 2 r ) ^ { 2 } - \pi r ^ { 2 } = 3 \pi r ^ { 2 } = 3 A _ { 1 }{/tex}
{tex} = 3 \times 346.5 \mathrm { cm } ^ { 2 } = 1039.5 \mathrm { cm } ^ { 2 }{/tex}
A₃ = Area of the region representing Blue scoring area ={tex} \pi ( 3 r ) ^ { 2 } - \pi ( 2 r ) ^ { 2 } = 9 \pi r ^ { 2 } - 4 \pi r ^ { 2 }{/tex}
{tex} = 5 \pi r ^ { 2 } = 5 A _ { 1 } = 5 \times 346.5 \mathrm { cm } ^ { 2 }{/tex}
{tex}= 1732.5 {/tex}cm²
 A₄ = Area of the region representing Black scoring area = {tex} \pi ( 4 r ) ^ { 2 } - \pi ( 3 r ) ^ { 2 } = 7 \pi r ^ { 2 } = 7 A _ { 1 }{/tex}
{tex} = 7 \times 346.5 \mathrm { cm } ^ { 2 }{/tex}{tex}= 2425.5{/tex} cm²
A₅ = Area of the region representing White scoring area = {tex} \pi ( 5 r ) ^ { 2 } - \pi ( 4 r ) ^ { 2 } = 9 \pi r ^ { 2 } = 9 A _ { 1 }{/tex}
{tex}= 9 \times 346.5{/tex} cm² {tex}= 3118.5{/tex} cm²