Question

Question 8 (ii)

The figure depicts a racing track. ADHE is a semicircular track of inner radius 30m and thickness 10m. FGBC is another semicircular track with the same inner radius and thickness as ADHE. The space between the two semicircular tracks are covered by straight tracks of length 106m and thickness 10m on either side. Calculate the area of the racing track.

Answer 1

Width of track = 10 m
Distance between two parallel lines = 60 m
Length of parallel tracks = 106 m

DE = CF = 60 m
Radius of inner semicircle, r = OD = O'C
$=\frac{60}{2}m=30m$
Radius of outer semicircle, R = OA = O'B
= 30 + 10 m = 40 m
Also, AB = CD = EF = GH = 106 m
Area of the track $=AreaofABCD+AreaEFGH+2\times \left(areaofoutersemicircle\right)-2\times \left(areaofinnersemicircle\right)$
$=(AB\times BC)+(EF\times FG)+2\times \left(\frac{\pi r^2}{2}\right)-2\times \left(\frac{\pi R^2}{2}\right)m^2=(106\times 10)+(106\times 10)+2\times \frac{\pi }{2}(r^2-R^2)m^2=2120+\frac{22}{7}\times 70\times 10m^2=4320m^2$