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.

Answer 1

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

= $\frac{1}{2}\times \pi \times ({49}^2-{35}^2)+90\times 14+\frac{1}{2}\times \pi \times ({49}^2-{35}^2)+90\times 14$
= $\pi \times (2401-12285)+2(90\times 14)$
= $\frac{22}{7}\times \left(1176\right)+2\left(1260\right)$

= 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

= $\frac{22}{7}\times 2\left(49\right)+180$

= 308 + 180

= 488 m