Find the inradius in a tangential quadrilateral

Question

In this tangential quadrilateral $AD+BC=22$, $AB=9$, $CF=8$ and $F$ is a tangent point. Find the inradius $r$.

Here is my attemps:

Using Pitot's theorem I found that $x+y=9$. Also, using the area formula for tangential quadrilateral i get that: $$ A=22\cdot r $$ Also, using the inradius formula in terms of the tangent segments I found that: $$ r^2=\frac{360+13\cdot x\cdot y}{22} $$

I draw the problem in AutoCAD and get that $x\approx 3.44836392$ and $y \approx 5.55163608$. So $r \approx 5.26080188$.

I wonder if it is possible to find a natural expression for $r$. I don't know what property should I use.

Many thanks if you can help me.

Answer 1

Though evident from the diagram, the AB || CD is not explicitly specified. Assume such to match the height

$$EG^2=(2r)^2= (x+8)^2-(x-8)^2 = (y+5)^2 - (y-5)^2\tag1$$

which leads to $8x=5y$ and, along with $x+y=9$, to get $x=\frac{45}{13}$ and $y=\frac {72}{13}$. Then, from (1), the inradius is

$$r= 3\sqrt{\frac{40}{13}}$$

Answer 2

The problem as stated has infinite number of solutions, since $x $ can take on any value between $0$ and $9$. Therefore, as follows from your formula, $r^2$ can take on any value in the range: $$\frac {360}{22}<r^2\le \frac {360+13\cdot\frac {81}4}{22}. $$