Solution:
Given the composite figure below;
The figure is made up of a triangle and a trapezoid of the following side measures,
[tex]\begin{gathered} For\text{ the triangle:} \\ b_1=6\text{ in} \\ h_1=8\text{ in} \\ For\text{ the trapezoid:} \\ a=10\text{ in} \\ b_2=16\text{ in} \\ h_2=9\text{ in} \end{gathered}[/tex]
To find the total area, A, of the figure, the formula is
[tex]\begin{gathered} A=Area\text{ of a triangle}+Area\text{ of a Trapezoid} \\ A=\frac{1}{2}b_1h_1+\frac{1}{2}(a+b_2)h_2 \\ Where \\ b_1\text{ is the base length of the triangle} \\ h_1\text{ is the height of the triangle} \\ b_2\text{ is the side length of the trapezoid} \\ h_2\text{ is the height of the trapezoid} \end{gathered}[/tex]
Substituting the values of the variables into the formula above
[tex]\begin{gathered} A=\frac{1}{2}(6)(8)+\frac{1}{2}(10+16)(9) \\ A=24+117=141\text{ in}^2 \\ A=141\text{ in}^2 \end{gathered}[/tex]
Hence, the total area, A, of the figure is 141 square inches.
