Fig. depicts a racing track whose left and right ends are semicircular. Fig. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

(i) the distance around the track along its inner edge 

(ii) the area of the track.

 

From the given information we draw the figure as follows,

The radius of the inner semicircles HIJ and KLG (r₁) = 60/2 = 30 m

Radii of the outer semicircles BCD and EFA = 30 m $+$ 10 m = 40 m

JK = GH = 106 m

DJ = HB = 10 m

The distance around the track along its inner edge:

= GH $+$ arc HIJ $+$ JK $+$ arc KLG

= 106 $+$ 2$\mathrm{\pi }$r₁/2 $+$106 $+$ 2$\mathrm{\pi }$r₁/2

= 106 $+$ $\mathrm{\pi }$$\times$30 $+$106 $+$$\mathrm{\pi }$$\times$30

= 212 $+$ 1320/7

= (1484 $+$ 1320)/7

= 2804/7 m

Now, the Radius of semicircle BCD = Radius of semicircle EFA 

(r₂) = 30 m $+$ 10 = 40 m

Area of the track = Area of rectangle ABHG $+$ Area of rectangle KJDE $+$ (Area of semicircle BCD - Area of semicircle HIJ) $+$ (Area of semicircle EFA - Area of semicircle KLG)

= {(106 $\times$ 10) $+$ (106 $\times$ 10) $+$ [1/2π (40)² - 1/2$\mathrm{\pi }$ (30)²] $+$ [1/2$\mathrm{\pi }$(40)² - 1/2$\mathrm{\pi }$ (40)²]

= 1060 $+$1060 $+$ [$\frac{1}{2}$$\mathrm{\pi }$ (1600 - 900)] $+$ [$\frac{1}{2}$$\mathrm{\pi }$ (1600 - 900)] 

= 1060 $+$ 1060 $+$ $\mathrm{\pi }$/2$\times$700 $+$ $\mathrm{\pi }$/2$\times$700

= 2120 $+$ 700$\mathrm{\pi }$

= 2120 $+$ 700 × 22/7

= 2120 $+$ 2200

= 4320 m²