The area of the figure can be found by dividing the figure into 2 triangles and 1 rectangle
using the pythagorean find the base of the triangle to the right
[tex]\begin{gathered} 5^2=4.6^2+x^2 \\ x=\sqrt[]{5^2-4.6^2} \\ x=\sqrt[]{25-21.16} \\ x=\sqrt[]{3.84} \\ x=1.959m\approx2.0m \end{gathered}[/tex]find the base of the the triangle to the left
[tex]\begin{gathered} y=10.6-2-4 \\ y=4.6 \end{gathered}[/tex]find the area for ach figure
[tex]\begin{gathered} A_1=4.6\cdot\frac{2}{2}m^2 \\ A_1=4.6m^2 \\ A_2=4\cdot4.6m^2 \\ A_2=18.4m^2 \\ A_3=4.6\cdot\frac{4.6}{2}m^2 \\ A_3=10.58m^2 \end{gathered}[/tex]add the three areas together
[tex]\begin{gathered} A=(10.58+4.6+18.4)m^2 \\ A\approx33.58m^2 \end{gathered}[/tex]