Triangle area from triangle

Let the position coordinates of $A, B, C$ be $(b, c), (0,0),(a, 0)$ respectively.

Then those of $K, L$ are $(\frac{a+2b}{4},\frac{c}{2}),(\frac{2a+b}{4},\frac{c}{4})$ respectively.

Hence, area of $\triangle {BKL}$ is

$9=\frac{1}{2}|\frac{a+2b}{4}\times \frac{c}{4}+\frac{2a+b}{4}\times (-\frac{c}{2})|=\dfrac {3ac}{32}$

$\implies$ area of $\triangle {ABC}$ is

$\dfrac {1}{2}ac=\boxed {48}$ .