SOLUTION: please help!!! the robinson track is shpaed like a rectangle with a semicircle on each of the shorter sies. the distance around the track is 1/4 mile. the straightaway is twice

Algebra ->  Surface-area -> SOLUTION: please help!!! the robinson track is shpaed like a rectangle with a semicircle on each of the shorter sies. the distance around the track is 1/4 mile. the straightaway is twice       Log On


   

Question 149114: please help!!!
the robinson track is shpaed like a rectangle with a semicircle on each of the shorter sies. the distance around the track is 1/4 mile. the straightaway is twice as long as the width of the field. what is the area of the field enclosed by the track to the nearest square foot?
thank you sooo much !
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the straight part of the track s
Call the width w
Adding up the stright parts on each side, s+%2B+s+=+2s
The straightaway is twice the width
s+=+2w
w+=+%281%2F2%29%2As
The circular portions of the track are made up of
1/2 the circumference of a circle with diameter = w on
one end and another 1/2 of a circumference with diameter = w
on the other end of the track
If circumference = C,
%281%2F2%29%2AC+%2B+%281%2F2%29%2AC+=+C
also,
C+=+pi%2Aw
C+=+3.1416%2A%281%2F2%29%2As
C+=+1.571s
1.571s+%2B+2s+=+.25mi
1 mi = 5280 ft
3.571s+=+.25+%2A+5280
s+=+369.6 ft
The area of the track is made up of a rectangle s%2Aw
and 2 half-circles which add up to a whole circle,
area = pi%2A%28w%2F2%29%5E2
pi%2A%28s%2F4%29%5E2
%283.1416+%2F+16%29%2A%28369.6%29%5E2
.196%2A136604.16
26774.42
The area of the rectangular part is s%2Aw
369.6%2A%28369.6%2F2%29
68302.08
The total area is
26774.42+%2B+68302.08+=+95076.5
The area of the field to the nearest ft2 is 95,077 ft2
check my work- I could have made a mistake, but I hope
you get my method