Question 1004187
x and y dimensions, d diagonal.


{{{system(2x+2y=68,x^2+y^2=d^2)}}}


Let d=26.


{{{system(x+y=34,x^2+y^2=26^2)}}}


{{{y=34-x}}}
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{{{highlight_green(x^2+(34-x)^2=26^2)}}}--------this can basically be the equation you want.  You can simplify it and solve for x, and then use this to solve for y.


{{{x^2+34^2-68x+x^2-26^2=0}}}
{{{2x^2-68x+34^2-26^2=0}}}
{{{2x^2-68x+(34+26)(34-26)=0}}}
{{{2x^2-68x+(60)(8)=0}}}
{{{2x^2-68x+480=0}}}
{{{highlight_green(x^2-34x+240=0)}}}------now simplified, quadratic equation


Discriminant,  {{{34^2-4*240=196=14^2}}}.


{{{x=(34+- 14)/2}}}, which would show BOTH dimensions x AND y.


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Dimensions would be 10 and 24.
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