  Question

The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90 m, and the ends are semicircles. If the track is 14 m wide everywhere, find the area of the track. Also, find the length of the outer boundary of the track.

Solution Let the radius of the inner semi-circular ends = r m.  Inner perimeter of the track = 400 m  90 + πr + 90 + πr = 400 (Circumference of the semi-circle = πr) 2πr = 220 m  r = 35 m  Area of the track = Area of the ring AEHD + Area of rectangle ABFE + Area of ring BFGC + Area of rectangle CDHG = 12× π × (492−352)+90 × 14 + 12× π × (492−352)+90 × 14 = π × (2401−12285)+2(90 × 14) = 227× (1176)+2(1260) = 3696 + 2520 = 6216 sq. m. Length of the outer running track = EF + Length of arc FG + GH + Length is arc HE = 90 + [π × (35 + 14)] + 90 + [ π × (35 + 14)] = [2π × 49] + 180 =  227× 2 (49)+180 = 308 + 180 = 488 m MathematicsSecondary School Mathematics XStandard X

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