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Question

An athletic track 14 m wide consists of two straight sections 120 m long joining semi-circular ends whose inner radius is 35 m. Calculate the area of the track.

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Solution

We have

OB=35 m and AB=CD=14 m,OA=(35+14) m=49 m


Now, Area of the track = Area of rectangle ABCD +Area of rectangle EFGH + 2{Area of the semi-circle with radius 49m} - 2{ Area of the semi-Circle with radius 35m}

=(AD×AB)+(EH×EF)+2[12π(49)2]2[12×π(35)2]

=(14×120)+(14×120)+2{12×227×(49)2}2{12×227×(35)2}

={3360+227(49+35)(4935)}=7056 m2


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