Is finding the area of this rectangle impossible?

Question

One of my students gave this problem and I am feeling quite ashamed that I could not find an answer. It asks for the area of the pink rectangle and it says that the triangle ABC is a right angle triangle.

I have tried using triangle similarity to find a system of equations, but I could not find a solution.

I would venture that it has infinite solutions, given the informations. But I fell that I am missing something really obvious.

The only assumptions are:

AB = 12
EC = 7
ABC is a right angled triangle
The pink shape is a rectangle

Answer 1

The area is $84-49\tan\beta$.

Rotate and reflect the picture to look like this

Complete $ABC$ to a rectangle $ABCO$. Let $O$ be the origin of Cartesian coordinates, let $OA$ be the $x$-axis, let $OC$ be the $y$-axis. The dimensions given show that point $E$ is $(7,12)$. Let $A = (a,0), P = (0,d)$. The two segments of the diagonal line have the same slope, that is $$ \frac{d}{a-7} = \tan\beta = \frac{12-d}{7} $$ Solve to get $$ d=12 - 7\tan\beta $$ The pink area is $$ (a-7)(12-d) = \frac{d}{\tan\beta}\cdot 7 \tan\beta = 7d = 84 - 49\tan \beta. $$

Answer 2

Suppose $\angle ECF=\alpha$. As can be seen in figure we have:

$AE=\frac{AB}{\tan \alpha}-EC$

$\tan \alpha=\frac{EF}{EC}\Rightarrow EF=\frac {CE}{\tan \alpha}$

area of rectangle is:

$S=[AE=(\frac{AB}{\tan \alpha}-CE)]( EF=CE\times \tan\alpha)=AB\times CE-CE^2\times\tan \alpha$

putting values of AB and CE we get:

$S=84-49\tan \alpha$

In particular case where $BE=2 AB$ we have:

$AE=\sqrt{24^2-12^2}=12\sqrt 3\Rightarrow AC=12\sqrt3+7\approx 27.76$

which gives :

$\tan\alpha=\frac{12}{27.76}\approx 0.4319\rightarrow \alpha\approx 23.36^o$

And area is:

$S=84-49\times 0.4319\approx 62.83$