Answer:
5 and one-third square feet.
Explanation:
Given a figure that is broken into a rectangle and a triangle.
The dimensions of the rectangle are:
[tex]\begin{gathered} Base=5\;ft \\ Height=\frac{1}{3}\;ft. \end{gathered}[/tex]The dimensions of the triangle are:
[tex]\begin{gathered} Base=3\frac{2}{3}\;ft. \\ Height=2\;ft. \end{gathered}[/tex]The area of the figure is calculated below:
[tex]\begin{gathered} Area=\text{ Area of Rectangle+Area of the triangle} \\ =LB+\frac{1}{2}bh \\ =(5\times\frac{1}{3})+(\frac{1}{2}\times3\frac{2}{3}\times2) \end{gathered}[/tex]The result is simplified below:
[tex]\begin{gathered} =\frac{5}{3}+\frac{11}{3} \\ =\frac{5+11}{3} \\ =\frac{16}{3} \\ =5\frac{1}{3}\;ft^2 \end{gathered}[/tex]The area of the figure is 5 and one-third square feet.
