Question

Fig. 12.26 depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segment 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

Answer 1

It is given that two Inner line

It is given that two inner line segment are $I{I}^{nd}$,

So, $EF\parallel HG$. Here $EF=AB=106m$

$HG=DC=106m$

$EH=FG=60m$

$AE=BF=HD=GC=10m$

distance around the track along inner edge = distance $EF$ + Distance $HG$ + Circumference of semicircle $EH$ with centre $O$ + Circumference of semicircle $FG$ with centre $O$.

Semicircle to $H$ with centre $O$

Diameter $=EH=60m$

So, radius $=\frac{Diameter}{2}=\frac{60}{2}=30$

Circumference of semicircle $EOH=\frac{1}{2}\times$ Circumference of circle

$=\frac{1}{2}\times \left(2\pi r\right)=\frac{1}{2}\times 2\times \frac{22}{7}\times 30$

So, circumference of semicircle $EOH=\frac{660}{7}m$

Similarly, As radius for both sei circles $EOH$ & $FOG$ are same circumference of semicircle $F{O}^{\prime }G$ = circumference of semi circle $EDH=\frac{660}{7}m$

Distance around the track along inner edge = distance $EF+$ distance $HG+$ circumference of semicircle $EH$ with centroid + circumference of semicircle $FG$ with centre ${}^{\prime }{O}^{\prime }$

$=106+106+\frac{660}{7}+\frac{660}{7}=212+\frac{660}{7}+\frac{660}{7}=\frac{10212+660+660}{7}$

$=\frac{2804}{7}c{m}^2$