Find area of triangle inscribed in a rectangle This is from an italian math competition.
In the rectangle below:

area of yellow, green and red triangles are 27, 35 and 40 resepectively. What is the area of the blue triangle ?
My trial: I tried to define $a,b,c$ and $d$:
$a(c+d)=80 \tag{1}$
$bc=70 \tag{2}$
$d(a+b)=54 \tag{3}$
(to clarify, a is the length of the side of the red triangle and b of the green triangle which lie on the top of the rectangle. c and d are the lengths of the sides of the green and yellow triangle lying on the right side of the rectangle )
This is a system of 3 equations in 4 unknowns. We can observe that if $(a,b,c,d)$ is a solution, so is $(ak,bk,c/k,d/k)$. We can therefore fix the value of one variable, $a=1$, solve the system, and obtain $b=5/4, c=56, d=24$.
The area of the entire rectangle would than be $R=(a+b)(c+d)=180$, and the blue area $78$.
This solution works, but I think it requires too many calculations for the type of competition. Is there a smarter/quicker way ?

It becomes simple if you see the trick. Let the height of the rectangle be $a$ and the width be $b$. Assign the shortest side of red $c$ and shortest side of yellow $d$. You will have, $d=\frac{54}{b}$ and $c=\frac{70}{a}$. Also you have, $(b-c)(a-d)=80 (1)$. Now, the area of triangle $A = ab - 102$.
If $x=ab$, substituting $c$ and $d$ is $(1)$ we get $(x-70)(x-54)=80x$. Solve the quadratic equation for $x$ and you will get $A$
