The rectangular WZYX is inside the rectangular ACDB
AB= 8 cm
AC= 6 cm
WX= 8 cm
What is WZ= ?
If and only if the points W, X, Y and Z has to be placed on the sides of the rectangle ABCD and $\displaystyle |\overline{WX}| = 8$ Then there is only one unique solution of this question. See attachment #2 to see what happens if $\displaystyle |\overline{AX}| $ is changing but $\displaystyle |\overline{WX}| $ remains constant. Only the thick outlined quadrilateral is a rectangle, all other quadrilaterals are trapezoids.
According to the labeling in attachment #1 you'll get:
$\displaystyle 8^2 = (8 - x)^2 + (6 - y)^2$ ........... [1]
The adjacent sides of the rectangle are perpendicular to each other. Using the slopes of the sides you'll get:
$\displaystyle -\frac yx \cdot \frac{6-y}{8-x}=-1$ ........... [2]
From [2] you'll get: $\displaystyle x = 4 - \sqrt{y^2 - 6·y + 16}$
Plug in the term of x into [1] and solve for y. I've to confess that this task was done by my calculator. You'll get:
$\displaystyle y \approx 1.895734320$ and consequently
$\displaystyle x \approx 1.133049932$
Since $\displaystyle |\overline{XY}| = |\overline{WZ}|$ you'll get $\displaystyle |\overline{WZ}| \approx \sqrt{1.895734320^2 + 1.133049932^2} \approx 2.208531358$
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