Answer:
[tex] 62.8 \: {m}^{2} [/tex]
Step-by-step explanation:
Radius of outer circle = 6 m
Width of the track = 2 m
So, radius of inner circle = 6 - 2 = 4 m
Area of the track = Area of outer circle - Area of inner circle
[tex] = \pi {6}^{2} - \pi {4}^{2} \\ \\ = \pi( {6}^{2} - {4}^{2} ) \\ \\ = 3.14 \times (36 - 16) \\ \\ = 3.14 \times 20 \\ \\ = 62.8 \: {m}^{2} [/tex]
