Question 1112853
For the quadrilateral in the sketch below, {{{d=6.74in}}} .
 
{{{drawing(400,325,-3.5,4.5,-3,3.5,
circle(0,0,2.5),
line(-2.815,-2.5,1.185,-2.5),
line(-2.815,-2.5,-2.137,3.2),
line(4.35,1.37,1.185,-2.5),
line(4.35,1.37,-2.137,3.2),
locate(-3.5,0.4,a=5in),
locate(-0.9,-2.5,b=4in),
locate(2.8,-0.6,c=5.75in),
locate(1.1,2.6,d),
circle(0,0,0.05),locate(0.7,0,green(r=2.5in)),
green(triangle(0,0,2.5,0,0,0))
)}}}
The Pitot theorem says that for a circumscribed quadrilateral with side lengths a, b, c and d in that order,
{{{a+c=b+d=(a+b+c+d)/2=s=semiperimeter}}} .
(In words, the sum of lengths of opposite sides is the semiperimeter).
Then, the area is  {{{K=r*s}}} where {{{r}}}=radius of incircle,
but {{{K<=sqrt(abcd)}}} .
The question did not state that the side length were given in order going around the quadrilateral,
but if we assume that another pair of sides are opposite,
the inequality above is not true,
meaning that a circumscribed quadrilateral side lengths could be
5, 4, 5.74, and 3.26, or
5, 4, 5.74, and 4.74, but then the circle would have to be smaller.