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Question
The given figure depicts a racing track whose left and right ends are semicircular.
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
[Use Π = 22/7]
Solution
Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA
`= 106 + 1/2xx 2pir + 106+ 1/2 xx 2pir`
`= 212 +1/2 xx2xx22/7xx30+1/2xx2xx22/7xx30`
`= 212 + 2 xx 22/7 xx 30`
`= 212 + 1320/7`
`=(1484+1320)/7 = 2804/7 m`
Area of the track = (Area of GHIJ − Area of ABCD) + (Area of semi-circle HKI − Area of semi-circle BEC) + (Area of semi-circle GLJ − Area of semi-circle AFD)
`= 106xx80-106xx60+1/2xx22/7xx(40)^2 - 1/2xx22/7xx(30)^2+1/2xx22/7xx(40)^2-1/2xx22/7xx(30)^2`
`=106(80-60)+22/7xx(40)^2-22/7 xx (30)^2`
`= 106(20)+22/7[(40)^2-(30)^2]`
`= 2120 + 22/7(40-30)(40+30)`
`=2120+(22/7)(10)(70)`
=2120+2200
= 4320 m2
Therefore, the area of the track is 4320 m2.
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