Find the radius of the circle. 
Let $ABCD$ be a quadrilateral in which $AB$ is parallel to $CD$ and perpendicular to $AD$;$AB$=3$CD$; and the area of the quadrilateral is 4. If a circle can be drawn touching all the sides of the quadrilateral, find its radius.

This is my problem .I found that $CD\cdot BD $=$2$ using the area of the quadrilateral. But I failed to find the radius. Actually I can't define or constructs the circle. Somebody please help me.
 A: 
For convenience call set $CD = t$.
Your quadrilateral is a trapezium with $t$ and $3t$ as the lengths of the parallel sides. The area is calculated by:
$A = \dfrac{t+3t}2 \cdot h = 4$ which gives  $h = \dfrac2t$ and therefore  $r = \dfrac1t$.
The length of  $x = t-  \dfrac1t$ and $y = 3t- \dfrac1t$
Since  $FB = 2t$ you can calculate the sides of the right triangle $CFB$, and hence by Pythagoras Theorem:
$$\left(\dfrac2t\right)^2 + (2t)^2=\left(t-\dfrac1t+3t-\dfrac1t\right)^2$$
$$\dfrac4{t^2}+4t^2=16t^2-16+\dfrac4{t^2}~\implies~12t^2=16~\implies~\boxed{  t=\sqrt{\dfrac43}}$$
Now put $t$ in $r = \dfrac1t$ to get the radius of the circle.
A: Let  $CD=a,\ AD=2b.\ \ \therefore AB=3a$.

Using simple geometry, it can be shown that the angle marked $\frac{\theta}{2}$ is, in fact, half of the angle marked $\theta$.
We know that,
$$
tan\ \theta=\frac{2\ tan\ \frac{\theta}{2}}{1-tan^{2}\frac{\theta}{2}}
$$
From figure,
$$
tan\ \theta=\frac{2b}{2a}=\frac{b}{a}
$$
and
$$
tan\frac{\theta}{2}=\frac{a-b}{b}
$$
Using the fact that area of quadrilateral is $4$, it can be shown that $a\cdot b=1$.

Solving the two equations, you get $b=\frac{\sqrt{3}}{2}$.
EDIT: On OP's request.
Given area of quadrilateral is 4.
$$
\frac{1}{2}\cdot (AB+CD)\cdot AD=4
$$ $$
\frac{1}{2}\cdot (3a+a)\cdot 2b=4
$$ $$
\therefore a\cdot b=1
$$
