We have the next figure
Because we have an irregular figure we divide the figure into a rectangle and in a triangle
For the area of the rectangle
[tex]A=l\times w[/tex]l=length
w= width
In our case
l=19
w=16
we substitute the values
[tex]A=19\times16=304in^2[/tex]for the area of the triangle
[tex]A=\frac{b\times h}{2}[/tex]b= base
h=height
in our case
b=16in
h=30-19=11in
we substitute the values
[tex]\begin{gathered} A=\frac{16\times11}{2} \\ A=88in^2 \end{gathered}[/tex]Then we sum the areas of the figures and we obtain the total are
[tex]\begin{gathered} A=304+88 \\ A=392in^2 \end{gathered}[/tex]