In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm. 
In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm. Suppose the lines $AD$ and $BC$ intersect at right angles and the lines $AC$ and $BD$ when extended at point $Q$ form an angle of $45^\circ$. Compute the area of $ABCD$.

What I Tried :- Here is the picture :-
Now to find the area of $ABCD$, I just need to find its height, but I cannot find it.
I can see that $\Delta AOB \sim \Delta COD$. So :-
$$\frac{AB}{CD} = \frac{AO}{OD} = \frac{BO}{OC} = \frac{2}{5}$$
So I assumed $AO = 2x$ , $BO = 2y$ , $CO = 5y$ , $DO = 5x$. 
Now in $\Delta AOB$, by Pythagorean Theorem :- 
$AO^2 + OB^2 = AB^2$ 
$\rightarrow 4x^2 + 4y^2 = 16$ 
$\rightarrow x^2 + y^2 = 4$
Also $\Delta QAB \sim \Delta QDC$. So:-
$$\frac{QA}{AC} = \frac{QB}{BD}$$ 
I get $AC$ and $BD$ by Pythagorean Theorem again, which gives me :-
$$\frac{QA}{\sqrt{4x^2 + 25y^2}} = \frac{QB}{\sqrt{25x^2 + 4y^2}}$$
I don't know how to proceed next, as this result only gives me that $\left(\frac{QA}{OB}\right)^2 = \frac{21y^2 + 16}{21x^2 + 16}$ . Also I couldn't think of any way to use the $45^\circ$ angle, except that I can figure out that the triangle is cyclic.
Can anyone help?

Let $OC = a$, $OD = b$. So $OA=\frac{2}{5}OC$, $OB = \frac{2}{5} OD$.
(Note you have swapped labels $C$ and $D$ in figure)
Also let $AD=3x$, $BC=3y$, so that $QA=2x$, $QB=2y$.
We have $a^2+b^2=100$
By Pythagoras,
$$ (OA^2+OD^2)+(OB^2+OC^2)=9(x^2+y^2) $$
$$ \Rightarrow x^2+y^2=116/9 $$
By cosine-rule in $\triangle QAB$,
$$ 4x^2+4y^2-4\sqrt{2}xy=4^2 $$
$$\Rightarrow xy=\dfrac{40\sqrt{2}}{9}$$
So
$$
\begin{align}
[ABCD] &= (1-\dfrac{4}{25})[QDC] \\
 &=\dfrac{21}{25}(\frac{1}{2}\cdot QD\cdot QC\cdot\sin 45^{\circ})\\ 
 &=\dfrac{21}{25}(\frac{1}{2}\cdot 5x\cdot5y\cdot\frac{1}{\sqrt{2}}) \\ 
 &=\boxed{\dfrac{140}{3}}
\end{align}
$$
