Question 1194281
x, y, the dimensions

{{{system(xy=sqrt(3),sqrt(x^2+y^2)=2)}}}

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I have not tried to solve this thoroughly, but ...

y^2+x^2=4
y^2=4-x^2



x^2*y^2=3
therefore
x^2*(4-x^2)=3,  one equation in just one unknown variable



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{{{-x^4+4x^2-3=0}}}
{{{x^4-4x^2+3=0}}}

{{{u=x^2}}}
THEN
{{{u^2-4u+3=0}}}
{{{(u-1)(u-3)=0}}}
{{{u=1}}} OR {{{u=3}}}


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