An athletic track 14 m wide consists of two straight sections 120 m long joining and semicircular ends inner diameter 35 cm . find area

Explanation

Given: The radius of the inner semi-circle, r=35 m and Width of the track =14 m

∴ Radius of the outer semi-circle, `R=35+14=49 m`

Length and breadth of rectangular track is l=120 m and b=14 m respectively

Area of the rectangular tracks= `2 xx (l xx b) = 2 xx (120 xx 14)= 3360 m^2`

Area of the semicircular tracks `=2 xx` (Area of the outer semicircle - Area of the inner semicircle)

Area of the semicircular tracks `=2 xx ((1/2) xx pi xx R^2 - (1/2) xx pi xx r^2) `

`:.`Area of the semicircular tracks `=2 xx (1/2) xx pi xx (R^2 - r^2)`

`:.`Area of the semicircular tracks `=2 xx (1/2) xx (22/7)xx ((49)^2 - (35)^2) =3696 m^2`

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