Question 935684
The diagonal of the rectangle is a hypotenuse for each of the two triangles that it forms.  


x and y, length and "width",
{{{system(x^2+y^2=7^2,xy=16)}}}


{{{y=16/x}}}, substitute this.
{{{x^2+(16/x)^2=49}}}
{{{x^4+16=49x^2}}}
{{{x^4-49x^2+16=0}}}, quadratic form.


{{{x^2=(49+sqrt(49^2-4*16))/2}}}
{{{x^2=(49+sqrt(2337))/2}}}

2337=3*779=3*19*41

Computing into base-ten values,
{{{highlight(x=sqrt((49+sqrt(2337))/2))}}}
x=0.328736 and y=48.671264
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Perimeter would be  {{{highlight(2(0.328736+48.671264))}}}, and you can choose what accuracy you want, or recalculate x and y to greater accuracy and use that if you want.



The quadratic form equation found in x would be the same if found in y, so the PLUS_OR_MINUS forms of the general solution method will give separately the x and the y.  Both are found and understood as positive values.