Solving for area of first figure
Recall the following formula for area of 2D figures
[tex]\begin{gathered} A_{\text{square}}=s^2 \\ A_{\text{trapezoid}}=\frac{a+b}{2}h \end{gathered}[/tex]The first figure consist of 2 figures with a square of side length of 2.5 cm, and a trapezoid with length 2.5 cm for the upper base, 5 cm for the lower base, and 2 cm for the height.
Calculate the area by getting the sum of the areas of the two figures
[tex]\begin{gathered} \text{Square: }s=2.5\text{ cm} \\ \text{Trapezoid: }a=2.5\text{ cm},b=5\text{ cm},h=2\text{ cm} \\ \\ A=s^2+\frac{a+b}{2}h \\ A=(2.5\text{ cm})^2+\frac{2.5\text{ cm}+5\text{ cm}}{2}\cdot2 \\ A=6.25\text{ cm}^2+\frac{7.5\text{ cm}}{\cancel{2}}\cdot\cancel{2}\text{ cm} \\ A=13.75\text{ cm}^2 \end{gathered}[/tex]The area of the first figure therefore is 13.75 square centimeters.
Solving for the area of the second figure
Recall the following areas for 2D figures
[tex]\begin{gathered} A_{\text{triangle}}=\frac{1}{2}bh \\ A_{\text{rhombus}}=bh \end{gathered}[/tex]Using the same procedures as above, we get the following
[tex]\begin{gathered} \text{Triangle: }b=2.5\text{ cm},h=2.2\text{ cm} \\ \text{Rhombus: }b=2.5\text{ cm},h=2\text{ cm} \\ \\ A=\frac{1}{2}(2.5\text{ cm})(2.2\text{ cm})+(2.5\text{ cm})(2\text{ cm}) \\ A=\frac{1}{2}(5.5\text{ cm}^2)+5\text{ cm}^2 \\ A=2.75\text{ cm}^2+5\text{ cm}^2 \\ A=7.75\text{ cm}^2 \end{gathered}[/tex]Therefore, the area of the second figure is 7.75 square centimeters.
