Check the picture below.
so we really have two identical trapezoids with bases of 2 and 6 and a height of 4, so let's get their area and sum them up and that's the area of the figure.
[tex]\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel ~sides}{bases}\\[-0.5em] \hrulefill\\ h=4\\ a=2\\ b=6 \end{cases}\implies \begin{array}{llll} A=\cfrac{4(2+6)}{2}\implies A=16\\\\ \stackrel{\textit{area for both trapezoids}}{2A=32} \end{array}[/tex]
