C is the midpoint of AD

\(\Rightarrow AC=CD=\dfrac{AD}{2}=5\left(unit\right)\)

\(\Rightarrow S_{\Delta CDE}=\dfrac{1}{2}\cdot CE\cdot5=30\left(units^2\right)\)

\(\Rightarrow CE=30\div5\div\dfrac{1}{2}=12\left(units\right)\)

\(\Rightarrow BC=\dfrac{2}{3}\cdot12=8\left(units\right)\)

\(\Delta ACB\) has \(\widehat{ACB}=\widehat{ECD}=90^o\left(\text{two opposite top angles}\right)\) 

=> \(\Delta ACB\) right at C

=> \(AB=\sqrt{8^2+5^2}=\sqrt{89}\left(units\right)\)

So AB = \(\sqrt{89}\) units