Not Pythagoras Time

Let $P$ and $Q$ is the midpoint of $AB$ and $CD$

Now let's find $AQ^2$ ,

$AQ^2 = \frac{1}{2} AC^2 + \frac{1}{2} AD^2 - \frac{1}{4} CD^2$

$AQ^2 = \frac{1}{2} (8)^2 + \frac{1}{2} (6)^2 - \frac{1}{4} (4)^2$

$AQ^2 = 32 + 18 - 4$

$AQ^2 = 46$

Now find $BQ^2$ ,

$BQ^2 = \frac{1}{2} BC^2 + \frac{1}{2} BD^2 - \frac{1}{4} CD^2$

$BQ^2 = \frac{1}{2} (7)^2 + \frac{1}{2} (5)^2 - \frac{1}{4} (4)^2$

$BQ^2 = \frac{49}{2} + \frac{25}{2} - 4$

$BQ^2 = 33$

Now let's find $PQ^2$

$PQ^2 = \frac{1}{2} AQ^2 + \frac{1}{2} BQ^2 - \frac{1}{4} AB^2$

$PQ^2 = \frac{1}{2} (46) + \frac{1}{2} (33) - \frac{1}{4} (3)^2$

$PQ^2 = 23 + \frac{33}{2} - \frac{9}{4}$

$PQ^2 = 37 \frac{1}{4}$

So $PQ^2 = n^2 = 37 \frac{1}{4} = \boxed{37.25}$