$ABCD$ is a trapezium where $AB$ || $CD.$ Let $AB = b$ , $CD = a$ where $a < b$. Let $S$ be the area of trapezium $ABCD$. It is given that $[\Delta BOC] = \frac{2S}{9}$. Find the value of $\frac{b}{a}$.
Related Question :- $ABCD$ is a trapezium where $AB$ $||$ $CD$. Suppose $[\Delta BOC] = \frac{25}{9}$, and let $AB = b$, $CD = a$ where $a < b$.
Note:- The real fact is that I copied the initial question wrong, saying that $[\Delta BOC] = \frac{25}{9}$ and there were no unique trapeziums of that kind, so here is the real question and sorry for the inconvenience.
What I Tried: Here is a picture :-
$ [\Delta BOC] = [\Delta AOD]$ follows from carpet strategy, so :-
$$ [\Delta DOC] + [\Delta AOB] = \frac{5S}{9}$$
We have that $\Delta DOC \sim \Delta BOA$ , which might help find $\frac{b}{a}$, but I couldn't understand how I can use it. One way is to put variables for $DO$ , $OC$ , $AO$ and $OB$ , but that will make it more complicated and I don't want to do it.
Another thing I thought of was to draw the perpendicular from $O$ , touching $AB$ and $CD$ at $M$ and $N$ , suppose. And let $[\Delta DOC] = x$ . I will get :-
$$\rightarrow \frac{1}{2} * ON * a = x$$
$$\rightarrow \frac{1}{2} * OM * b = \frac{5S}{9} – x$$
But that still deals with a variable $x$ , how do I fix it?
Can anyone help me? Thank You.
Best Answer
$$\frac{DO}{OB}=\frac{ar(\Delta DOC)}{2S/9}=\frac{a}{b}={\left(\frac{ar(\Delta DOC)}{ar(AOB)}\right)}^{1/2}$$
let $ar(DOC)=p,ar(AOB)=q$ then by above $$pq=\frac{4S^2}{81}$$ then use $$p+q+4S/9=S$$ from here $p=?,q=?\Rightarrow \frac{a}{b}=?$