Think Geometry!

We note that $\triangle BCD$ is similar to $\triangle ABC$ . And $\triangle ABC$ is a 3-4-5 Pythagorean triangle.

$\implies \begin{cases} BD = \dfrac 45 BC \\ CD = \dfrac 35 BC \end{cases} \implies BD+CD = \left( \dfrac 45+ \dfrac 35\right) BC = \dfrac 75 \times 6 = 8.4$

$\implies \lfloor 100k \rfloor = \boxed{840}$

$Right\ \Delta\ ABC\ is\ \ 3-4-5,\\ \implies\ right\ \Delta\ CDB\ is\ also\ 3-4-5.\\ But\ BC\ is \ hypotenuse\ of\ \Delta\ CDB,\ DC \ the\ small\ side.\\ \therefore\ BC=6*\frac 5 5,\ \ \ BD=6*\frac 4 5\ \ \ DC=6*\frac 3 5.\\ \implies\ k=6*\frac 4 5+6*\frac 3 5=6*\frac {4+3} 5=8.4.\\ \lfloor 100k \rfloor=840.$

$BD = \frac{AB \times BC}{AC}$

$= 4.8$ units.

$BC²= CD \times AC$

$CD = 3.6$ units

$BC + CD = 8.4 = k$ units

$\lfloor 100k \rfloor = 840.$