Let
A₁ be the area of the parallelogram
A₂ be the area of the trapezoid
Solving for the area of the parallelogram
Given the following dimensions
b = 23 cm
h = 14 cm
The area is solved using
[tex]\begin{gathered} A_1=bh \\ A_1=(23\text{ cm})(14\text{ cm}) \\ A_1=322\text{ cm}^2 \end{gathered}[/tex]
The area of the parallelogram therefore is 322 square centimeters.
Solving for the area of the trapezoid.
Given the following dimensions
b₁ = 15 cm
b₂ = 34 cm
h = 19 cm
The area is solved using
[tex]\begin{gathered} A_2=\frac{b_1+b_2}{2}\cdot h \\ A_2=\frac{15\text{ cm}+34\text{ cm}}{2}(19\text{ cm}) \\ A_2=\frac{49\text{ cm}}{2}(19\text{ cm\rparen} \\ A_2=(24.5\text{ cm})(19\text{ cm}) \\ A_2=465.5\text{ cm}^2 \end{gathered}[/tex]
The area of the trapezoid is 465.5 square centimeters.
Solving for the area of the composite figure.
Get the sum of the two areas to get the area of the composite figure, we have
[tex]\begin{gathered} A_{\text{total}}=A_1+A_2 \\ A_{\text{total}}=322\text{ cm}^2+465.5\text{ cm}^2 \\ A_{\text{total}}=787.5\text{ cm}^2 \end{gathered}[/tex]
Therefore, the area of the composite figure is 787.5 square centimeters.
