4300 m²
1) As this is not a regular pentagon for their lengths are not congruent to each other let's decompose that pentagon into other polygons:
So we have a quadrilateral below and a triangle on top.
Area of that triangle below:
[tex]A=\frac{1}{2}\cdot b\cdot h\rightarrow A=100\times60\times\frac{1}{2}=300m^2[/tex]
Considering that quadrilateral is a parallelogram with two pairs of parallel and congruent sides
[tex]\begin{gathered} A=w\cdot l \\ A=80\times50 \\ A=4000m^2 \end{gathered}[/tex]
The area of that pentagon:
[tex]A=4000+300,A=4300m^2[/tex]
