Given :
To Find :
Solution :
Area of the outer circle :
[tex]\: \qquad \dashrightarrow \sf{{ \pi \: \times {(490)}^{2} {m}^{2}}} [/tex]
[tex]\: \qquad \dashrightarrow \sf{{ \ \dfrac{22}{7} \times \:490 \times 490 \: {m}^{2}}} [/tex]
[tex]\: \qquad \dashrightarrow \sf{{ \ \dfrac{22}{ \cancel{7}} \times \cancel{490} \times 490 \: {m}^{2}}} [/tex]
[tex]\: \qquad \dashrightarrow \sf{{ \ {22} \times 70 \times 490 \: {m}^{2}}} [/tex]
[tex]\: \qquad \dashrightarrow \bf{{ \ 754600 \: {m}^{2}}} [/tex]
⠀
Area of the inner circle :
[tex]\: \qquad \dashrightarrow \sf{{ \pi \: \times {(210)}^{2} {m}^{2}}} [/tex]
[tex]\: \qquad \dashrightarrow \sf{{ \ \dfrac{22}{7} \times \:210 \times 210 \: {m}^{2}}} [/tex]
[tex]\: \qquad \dashrightarrow \sf{{ \ \dfrac{22}{ \cancel{7}} \times \cancel{210} \times 210 \: {m}^{2}}} [/tex]
[tex]\: \qquad \dashrightarrow \sf{{ \ {22} \times 30 \times 210 \: {m}^{2}}} [/tex]
[tex]\: \qquad \dashrightarrow \bf{{ \ 138600 \: {m}^{2}}} [/tex]
⠀
Therefore,
Area of the region inside the circular paths :
[tex]\: \qquad \dashrightarrow \sf{{ \ (754600 - 138600 ) \: {m}^{2}}}[/tex]
[tex]\: \qquad \dashrightarrow \bf{{ \ 616000 \: {m}^{2}}}[/tex]
