Question 149114
Call the straight part of the track {{{s}}}
Call the width {{{w}}}
Adding up the stright parts on each side, {{{s + s = 2s}}}
The straightaway is twice the width
{{{s = 2w}}}
{{{w = (1/2)*s}}}
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}}},
{{{(1/2)*C + (1/2)*C = C}}}
also,
{{{C = pi*w}}}
{{{C = 3.1416*(1/2)*s}}}
{{{C = 1.571s}}}
{{{1.571s + 2s = .25}}}mi
1 mi = 5280 ft
{{{3.571s = .25 * 5280}}}
{{{s = 369.6}}} ft
The area of the track is made up of a rectangle {{{s*w}}}
and 2 half-circles which add up to a whole circle,
area = {{{pi*(w/2)^2}}}
{{{pi*(s/4)^2}}}
{{{(3.1416 / 16)*(369.6)^2}}}
{{{.196*136604.16}}}
{{{26774.42}}}
The area of the rectangular part is {{{s*w}}}
{{{369.6*(369.6/2)}}}
{{{68302.08}}}
The total area is
{{{26774.42 + 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