Intersection of right triangles

Be $F$ the point of intersection of the segments $AB$ and $DE$ . The $\triangle ABC$ is similar to the $\triangle AEF$ . We can get this relation:

The segment $AE$ has a length of $a$ , and the segment $EF$ has a length of $x$ . We need to find the value of $x$ in terms of $a$ or $b$ . How the triangles is similar the following relationship is valid:

$\frac{BC}{AC}=\frac{EF}{AE}$ and with it we can find $x$ :

$\frac{a}{b}=\frac{x}{a}$

$bx=a^2$

$x=\frac{a^2}{b}$

The area of the blue triangle is $\frac{EF\cdot AE}{2}$ . Then, $\frac{\frac{a^2}{b}\cdot a}{2}=\frac{a^3}{2b}=a^3\cdot b^{-1} \cdot \frac{1}{2}$ . The answer must be $3\cdot (-1)\cdot\frac{1}{2}=-1.5$