The area is 58^.5

Rotate $~$ $\triangle APD$ $~$ to $~$ $\triangle AQB$ $~$ by ($\pi$/2). $\angle PAQ=$ ($\pi$/2) , $PQ=3\sqrt{2}$ and apply 2nd Cosine rule to $~$ $\triangle PQB$ $~$ $\Longrightarrow$ $~$ $\angle QPB=$ ($\pi$/4). Thus, $~$ $\angle APB=$ ($\pi$/2) and then $AB^{2}=AP^{2}+BP^{2}=58$

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