Question 1150827
the diagonal of a rectangle is {{{d=15cm}}}, 

recall: 

{{{d^2=a^2+b^2}}} where {{{a}}} and {{{b}}} are sides of the rectangle

so, {{{15^2=a^2+b^2}}}

{{{225=a^2+b^2}}}..........eq.1

 
and, if the perimeter is {{{P=38}}}, means {{{2(a+b)=38}}}

{{{a+b=38/2}}}

{{{a+b=19}}}.......solve for {{{a}}}

{{{a=19-b}}}..........eq.2

go to

{{{225=a^2+b^2}}}..........eq.1...substitute {{{a}}}

{{{225=(19-b)^2+b^2}}}.........solve for {{{b}}}

{{{225=19^2-38b+b^2+b^2}}}

{{{225=361-38b+2b^2}}}...swap the sides

{{{2b^2-38b+361=225}}}

{{{2b^2-38b+361-225=0}}}

{{{2b^2-38b+136=0}}}....both sides divide by {{{2}}}

{{{b^2-19b+68=0}}}.......use quadratic formula

 {{{b = (-(-19) +- sqrt( (-19)^2-4*1*68 ))/(2*1) }}}

{{{b = (19 +- sqrt( 361-272 ))/2 }}}

{{{b = (19 +- sqrt( 89 ))/2 }}}.........since we are looking for side length, we need only positive root

{{{b = (19 +sqrt( 89 ))/2 }}}-> exact solution

{{{b = 14.216991}}}-> approximately


now find {{{a}}}


{{{a=19-b}}}..........eq.2

{{{a=19-(19 +sqrt( 89 ))/2}}}

{{{a=(38-19 -sqrt( 89 ))/2}}}

{{{a=(19 -sqrt( 89 ))/2}}}-> exact solution

{{{a=4.78301}}}-> approximately


 then,  the area will be:

{{{A=a*b}}}......using exact solutions we have

{{{A=((19 -sqrt( 89 ))/2)*((19 +sqrt( 89 ))/2)}}}

{{{A=68cm^2}}}

or, using approximate solutions

{{{A=a*b}}}

{{{A=4.78301*14.216991}}}

{{{A=68.00001012291}}}.....rounded it is

{{{A=68cm^2}}}