hello
to solve this question, we need to solve each object separately. i.e the triangle and the rectangle separately.
[tex]\begin{gathered} \text{area of the triangle =}\frac{1}{2}b\times h \\ \text{area of the rectangle = l}\times w \end{gathered}[/tex][tex]\begin{gathered} \text{width(w)}=\frac{3}{2}=1.5ft \\ \text{height}=6ft \\ \text{area of the triangle = 2}\times\text{(}\frac{1}{2}\times1.5\times6) \\ \text{area of the triangle= 9ft}^2 \end{gathered}[/tex]
area of the rectangle is
[tex]\begin{gathered} a=l\times w \\ l=8ft \\ w=16ft \\ a=8\times16 \\ a=128ft^2 \end{gathered}[/tex]
now the total area of the irregular object is equal to the sum of the area of the triangle + area of the rectangle
[tex]\begin{gathered} \text{area}=9+128 \\ \text{area}=137ft^2 \end{gathered}[/tex]
from the calculation above, the area of the irregular shape is 137 squared feet
