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put this solution on YOUR website!FIND THE WIDTH OF THE FIELD (X), THE LENGTH OF THE STRAIGHTAWAY (2X) ANS THE
CIRCUMFERANCE OF THE SEMICIRCLES (2PIR) AND THE AREA OF THE CIRCLE (PIR^2).
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STRAIGHT AWAY=2*2X=4X
END SEMICIRCLES=2PIX=2*3.14X=6.28X
4X+6.28X=5,280
10.28X=5,280
X=5,280/10.28
X=513.62 IS THE WIDTH OF THE FIELD.
2*513.62=1,027.24 LENGTH OF THE STRAIGHTAWAY.
2*3.14*513.62=6.28*513.62=3,225.53 DISTANCE AROUND THE 2 SEMI CIRCLES.
PROOF:
2*1,027.24+3,225.53=2054.48+3,225.53=5,280 DISTANCE OF THE TRACK.
3.14*513.62^2=3.14*263,805.5 AREA OF THE END SEMI CIRCLES.
496.8*939.6=466,793.28 AREA OF THE RECTANGLE.
466,793.28+233,396.64=700,189.92 FT^2 ANS. FOR THE AREA OF THE FIELD.