Question

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.

Answer 1

Distance around the track along its inner edge =AB +arc BEC +CD +arc DFA
$=106+\frac{1}{2}\times 2\pi r+106+\frac{1}{2}\times 2\pi r=212+\frac{1}{2}\times 2\times \frac{22}{7}\times 30+\frac{1}{2}\times 2\times \frac{22}{7}\times 30=212+2\times \frac{22}{7}\times 30=212+\frac{1320}{7}=\frac{1484+1320}{7}=\frac{2804}{7}m$
Area of the track =(Area of GHIJ-Area of ABCD) +(Area of semi -circle HKl-Area of semi -circle GLI -Area of semi -circle AFD)
$=106\times 80-106\times 60+\frac{1}{2}\times \frac{22}{7}\times (40{)}^2-\frac{1}{2}\times \frac{22}{7}\times (30{)}^2+\frac{1}{2}\times \frac{22}{7}\times (40{)}^2-\frac{1}{2}\times \frac{22}{7}\times (30{)}^2=106(80-60)+\frac{22}{7}\times (40{)}^2-\frac{22}{7}\times (30{)}^2=106\left(20\right)+\frac{22}{7}\left[\right(40{)}^2-\left(30{)}^2\right]=2120+\frac{22}{7}(40-30)(40+30)=2120+\left(\frac{22}{7}\right)\left(10\right)\left(70\right)=2120+2200=4320m^2$
Therefore, the area of the track is $4320m^2$.