check the picture below.
so, then using the pythagorean theorem, we can get "b", namely the height of the trapezoid.
[tex]\bf \textit{area of a trapezoid}\\\\
A=\cfrac{h(a+b)}{2}~~
\begin{cases}
a,b=\stackrel{bases}{parallel~sides}\\
h=height\\
---------\\
a=3\\
b=\stackrel{1+4}{5}\\
h=\sqrt{8}
\end{cases}\implies A=\cfrac{\sqrt{8}(3+5)}{2}
\\\\\\
A=\cfrac{8\sqrt{8}}{2}~~
\begin{cases}
8=2\cdot 2\cdot 2\\
\qquad 2^2\cdot 2
\end{cases}\implies A=\cfrac{8\sqrt{2^2\cdot 2}}{2}\implies A=\cfrac{8\cdot 2\sqrt{2}}{2}
\\\\\\
A=\cfrac{16\sqrt{2}}{2}\implies A=8\sqrt{2}[/tex]
