Question 779431

rectangle with sides {{{a}}} and {{{b}}} with a semicircle at each end tells you that diameter {{{d}}} of a semicircle is equal to the one side of the rectangle; let it be side {{{b}}}

then the radius of a semicircle is {{{r=b/2}}}

The perimeter of the entire track is the perimeter of 2 semicircles  plus two sides of the rectangle {{{2a }}} (we do not need diameters {{{d}}} or sides {{{b}}}, they are inside a track).

since they are congruent, the perimeter of 2 semicircle equal to the perimeter of a circle which is {{{2rpi}}} or {{{2(b/2)pi=b*pi}}} since {{{r=b/2}}}

then

 {{{P= 2a+b*pi=200pi* ft}}}......equation 1

the area of only the rectangular region is 

{{{A=ab=5000pi* ft^2}}}......equation 2

solve this system:

{{{2a+b*pi=200pi* ft}}}......equation 1

{{{ab=5000*pi* ft^2}}}......equation 2....solve for {{{a}}}
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{{{a=(5000pi)/b}}}......equation 2.....plug in equation 1


{{{2(5000pi)/b+b*pi=200pi}}}......equation 1...solve for {{{b}}} and cancel {{{pi}}}


{{{(10000)/b+b =200}}}.....multiply by {{{b}}}


{{{cross(b)(10000)/cross(b)+b^2=200b}}}


{{{10000+b^2 -200b=0}}}


{{{b^2 -200b+10000=0}}}


{{{(b -100)^2=0}}}


{{{b -100=0}}}


{{{b =100ft}}}..........now we know radius {{{r=50}}}


the width of the rectangular region is {{{b =100ft}}} 

the length of the rectangular region will be

{{{a=(5000pi)/100}}}

{{{a=50pi*ft}}}