Length of a line

$\angle EBA=180-\angle ABC$ and $\angle ADF=180-\angle ADC$

However, $\angle ABC=\angle ADC$ . Therefore, $\angle EBA=\angle ADF$ .

It follows that, $\triangle EBA \sim \triangle ADF$ $(A.A.)$ . So we have

$\dfrac{DF}{AD}=\dfrac{EB}{AB}$ $\color{#D61F06}\large \implies$ $\dfrac{DF}{5}=\dfrac{3}{4}$ $\color{#D61F06}\large \implies$ $DF=\dfrac{3(5)}{4}=$ $\color{#3D99F6}\boxed{\large 3.75}$

For calculating the area if ABCD we have 2 options:

$AD \cdot AE$ and $AB \cdot AF$ .

Obviously these give the same result. Using Pythagoras we can find $AE = \sqrt7$ . So we have:

$5 \cdot \sqrt7= 4 \cdot AF$ .

So $AF = \frac{5}{4}\sqrt7$ . Now using Pythagoras in triangle ADF we can calculate $DF= 3.75$ .