trapezoid

$\tan~\theta = \dfrac{3}{8}$ $\implies$ $\theta \approx 22.556$

$\cos~20.556 = \dfrac{3}{x}$ $\implies$ $x=\dfrac{3}{\cos~20.556}$ $\implies$ $x \approx 3.204$

By similar figures, we have

$\dfrac{3}{8}=\dfrac{y}{5}$ $\implies$ $y=\dfrac{3(5)}{8}\dfrac{15}{8}=1.875$

Let $A_B$ be the area of the big trapezoid and $A_S$ be the area of the small trapezoid. Then

$A_B=\dfrac{1}{2}(14+20)(8)=136$

$A_S=\dfrac{1}{2}(9.842+13.592)(5)\approx 58.585$

Finally the area of the figure is $136-58.585 \approx$ $\boxed{77}$