so we have a trapezoid with a parallelogram inside.
now, if we just get the area of the trapezoid, which includes the parallelogram, and then get the area of the parallelogram and subtract it from that of the trapezoid, what's leftover is the shaded region, because we'd be in effect making a "hole" in the trapezoid and the area leftover is the shaded part.
[tex]\bf \stackrel{\textit{area of a trapezoid}}{A=\cfrac{h(a+b)}{2}}~~ \begin{cases} a,b=\stackrel{bases}{parallel}\\ \qquad ~~ sides\\ h=height\\ \cline{1-1} a=19\\ b=25\\ h=20 \end{cases}\qquad \stackrel{\textit{area of a parallelogram}}{A=bh}~~ \begin{cases} b=base\\ h=height\\ \cline{1-1} b=17\\ h=10 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{trapezoid}}{\cfrac{20(19+25)}{2}}~~-~~\stackrel{\textit{parallelogram}}{(17\cdot 10)}\implies 10(44)-170\implies 440-170\implies 270[/tex]
