Question

Figure depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is $60m$ and they are each $106m$ long. If the track is $10m$ wide, find:
(i) the distance around the track along its inner edge
(ii) the area of the track

Answer 1

Here, we have
$OE=O^{\prime }G=30m$
$AE=CG=10m$
$OA=O^{\prime }C=(30+10)m=40m$
$AC=EG=FH=BD=106m$
(i)
The distance around the track along its inner edge
$=EG+FH+2\times$(circumference of the semicircle of radius $OE=30cm$)
$=106+106+2(\frac{1}{2}\times 2\pi \times 30)=212+60\pi$
$=212+60\frac{22}{7}=(212+\frac{1320}{7})m=\left(\frac{1484+1320}{7}\right)m=\frac{2804}{7}m=400\frac{4}{7}m$
(ii)
Area of the track = Area of the shaded region
= Area of the rectangle $AEGC+$ Area of the rectangle $BFHD+2+$ (Area of the semicircle of radius $40m-$ Area of the semicircle with radius $30m-$)
$=\left[\right(10\times 106)+(10\times 106\left)\right]+2\{\frac{1}{2}\times \frac{22}{7}\times (40{)}^2-\frac{1}{2}\times \frac{22}{7}\times \left(30{)}^2\right\}$
$=1060+1060+\frac{22}{7}\left[\right(40{)}^2-\left(30{)}^2\right]$
$=2120+\frac{22}{7}\times 700=2120+2200=4320m^2$