A geometry problem by A Former Brilliant Member

Since $\triangle BAG \sim \triangle FDG$ , we have, $\dfrac{GF}{11}=\dfrac{4}{14}$ .

$GF=\dfrac{4(11)}{14}=\dfrac{22}{7}$

Since $\triangle EBC \sim \triangle EGA$ , we have, $\dfrac{BE}{EG}=\dfrac{10}{14}$ .

$\dfrac{BE}{11-BE}=\dfrac{10}{14}$ $\color{#D61F06}\implies$ $14BE=10(11-BE)$ $\color{#D61F06}\implies$ $24BE=110$ $\color{#D61F06}\implies$ $BE=\dfrac{55}{12}$

Finally,

$EF=11-BE-GF=11-\dfrac{55}{12}-\dfrac{22}{7}=\dfrac{275}{84}=$ $\boxed{3.2738}$