Determinate the shaded area of the rectangle divided in 4 triangles

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Determinate the area of the shaded area of the rectangle divided in $4$ triangles

The areas of the triangles are the numbers $3,4,5$ and with those areas we should get the area of the shaded triangle.

No trigonometry

My try: I made a very messy equation system but i concluded that my equation system is wrong because i got that the segments of the left side of the rectangle (apparently the shortest cathetus of the triangle with area $3$ and apparently the shortest cathetus of the triangle with area $5$) are the same (the image shows that's not right).

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Let the height of the rectangle be $x$ and the width $y$. Then

  • the upper side of the area $4$ triangle is $8/x$;
  • the upper side of the area $3$ triangle is $y-(8/x)$;
  • the left side of the area $3$ triangle is $\dfrac{6}{y-(8/x)}$.

So by looking at the triangle of area $5$ we have $$5=\frac12\,y\,\left(x-\frac{6}{y-(8/x)}\right)$$ which simplifies to $$(xy)^2-24(xy)+80=0$$ with solutions $xy=20,4$. Obviously $xy>12$ so reject $4$, so the rectangle has area $20$ and the triangle has area $8$.