1)  I took the approach that a square is also a trapezoid, so if the answer is true for every isosceles trapezoid which can have an inscribed circle, then it must also be true for a square with an inscribed circle.  I made it a unit circle inside a square of side 2.

So making an assumption which may or may not be valid:

AL = LC = (√2 + 1) = s

AK = LC = (√2 - 1) = d

So the expression becomes s^2 / (d^2)^(1/4)

(√2 + 1)^2 / sqrt(√2 - 1)

Multiply top and bottom by conjugate:

(√2 + 1)^2 * sqrt(√2 + 1) / 1

(3 + 2√2) * sqrt(√2 + 1)

sqrt((3 + 2√2)^2) * sqrt(√2 + 1)

sqrt( (17 + 12√2)*(√2 + 1) )

√( 41 + 29√2 ) =  approx  9.056

2) but wait ...  If you check the units, L for length, we have L^2 / L^(1/2)  or L^(3/2).  It is not a pure dimensionless number.  So if assuming the circle radius is 1 leads to 'answer', then assuming radius is 2 would multiply 'answer' by 2^1.5  

So I think the problem as stated, without defining the scale, does not have a single numeric answer, but must depend on, for example r, the radius of the circle.

So I will amend my answer to √( 41 + 29√2 ) * r^(3/2)