A 21m wide athletic track consists of two straight sections 150m long joining semi circular ends whose diameters are 84m each . Find the area of the track .
 

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}2\mathrm{straight}\mathrm{sections}=2\times 150\mathrm{m}\times 21\mathrm{m}=6300{\mathrm{m}}^2 \\ \mathrm{diameter}=84 \mathrm{m}, \mathrm{radius}=\frac{84}{2}=42 \mathrm{m} \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{inner}\mathrm{semicircle}=42\mathrm{m} \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{outer}\mathrm{semicircle}=42\mathrm{m}+21 \mathrm{m}=63\mathrm{m} \\ \mathrm{here} \mathrm{\pi }=\frac{22}{7} \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{semicircular}\mathrm{region}=\frac{1}{2}\times \mathrm{\pi }\left({63}^2-{42}^2\right){\mathrm{m}}^2=3268.5{\mathrm{m}}^2 \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{semicircular}\mathrm{regions}=2\times 3268.5{\mathrm{m}}^2=6537{\mathrm{m}}^2 \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{track}=\left(6300+6537\right){\mathrm{m}}^2=12837 {\mathrm{m}}^2 \end{gathered}$$

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