For this type of problem we add up the individual areas of each region, in this case we have 3 squares and 2 triangles, the squares have sides of 6 m of lenght and the triangles have a base of 6 m and a height of 4 m.
Recalling that the area of a square is sides* side, the area of each square is
![A_s=36m^2_{}](https://img.qammunity.org/2023/formulas/mathematics/college/86twytbq87sk59usqwk4yl2z44ijo1hof9.png)
and the area of a triangle is
![A_t=(bh)/(2)=(6m\cdot4m)/(2)=(24)/(2)m^2=12m^2](https://img.qammunity.org/2023/formulas/mathematics/college/dsqyox9bmhttpqkpxfie3yqir7pva0kwu1.png)
Finally we add up the areas:
![A=3A_s+2A_t=3(36m^2)+2(12m^2)=132m^2](https://img.qammunity.org/2023/formulas/mathematics/college/5x2j9dy0zewkkkkrm2fy5zlx6t8jmqbth5.png)
Answer: The area is 132m^2.