$ABCD$ is a trapezium where $AB$ || $CD.$ Let $AB = b$ , $CD = a$ where $a < b$. Let $S$ be the area of trapezium $ABCD$.

Question

$ABCD$ 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.

Answer 1

$$\frac{DO}{OB}=\frac{ar(\Delta DOC)}{2S/9}=\frac{a}{b}={\left(\frac{ar(\Delta DOC)}{ar(AOB)}\right)}^{1/2}$$

let $ar(DOC)=p,ar(AOB)=q$ then by above $$pq=\frac{4S^2}{81}$$ then use $$p+q+4S/9=S$$ from here $p=?,q=?\Rightarrow \frac{a}{b}=?$

Answer 2

I showed in my answer to a related post if $[COD]=P$, $[AOB]=Q$, then for trapezium $ABCD$, $$[BOD]=\sqrt{PQ}=[AOC] \quad , \quad [ABCD]=(\sqrt{P}+\sqrt{Q})^2$$

Given here is $$\dfrac{P}{Q}=\dfrac{a^2}{b^2} \quad , \quad \dfrac{\sqrt{PQ}}{(\sqrt{P}+\sqrt{Q})^2}=\dfrac{2}{9}$$

Dividing second one by $Q$, and letting $a/b=x$ we get $$\dfrac{x}{(1+x)^2}=\dfrac{2}{9}$$ which can be solved to obtain $$x= \dfrac{1}{2} \Rightarrow \dfrac{b}{a}=2$$

Answer 3

Using your results you can set up directly the following two equations using the notations:

• $\triangle DOC$ with area $\Delta_u$ and height $h_u$ and
• $\triangle ABO$ with area $\Delta_l$ and height $h_l$
• $h= h_l+h_u$ - height of the trapezium

So, because of similarity we have

$$r := \frac{h_l}{h_u}=\frac{b}{a}$$

Now, using your result, you have

$$\frac{5}{9}S = \Delta_u + \Delta_l =\Delta_u(1+r^2)$$ and $$S = \frac{a+b}{2}h = \frac{a+b}{2}(h_u+h_l)=\frac a2h_u(1+r)(1+r)=\Delta_u(1+r^2)$$

Divide the equations to get $$\frac 95 = \frac{(1+r)^2}{1+r^2}$$

This has two solutions for $r$ where the one in question is $\boxed{r=2}$, since $b>a$.

Answer 4

We connect midpoints of DC and AB, if passes O. We can easily see that $S_{EGBC}=S_{ADEG}= \frac S 2$ abd we have:

$S_{DOC}=S-4(\frac{2s}9)=\frac S 9$

$2 a \times h=\frac{4s}9+2 S_{DOC}=\frac{6S}9=\frac{2S}3$ ⇒ $a=\frac{2S}{3h}$

$S=(a+b)\frac h 2$ ⇒ $a\times \frac h 2 +b\frac h 2=S$ ⇒ $b=\frac{4 S}{3 h}$

⇒ $\frac b a = 2$