We are asked to find the area of the shaded region.
Notice that the overall figure is a rectangle and the missing area is a triangle.
The area of the rectangle is given by
![A_{\text{rect}}=w\cdot l](https://img.qammunity.org/2023/formulas/mathematics/high-school/9kcerrp4nav51aun16ud7ztbeusc39x51e.png)
Where w is the width and l is the length of the rectangle.
width = 7 m
length = 16 m
![A_{\text{rect}}=w\cdot l=7\cdot16=112\; \; m^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/d5azudikza0l09lezvrr02palgwy1gh65u.png)
So, the area of the rectangle is 112 m²
The area of the triangle is given by
![A_{\text{tri}}=(1)/(2)\cdot b\cdot h](https://img.qammunity.org/2023/formulas/mathematics/high-school/669an4xt8p6ho9zf4pxgqyt9amwkgg5dy8.png)
Where b is the base and h is the height of the triangle.
base = 7 m
height = 8 m
![A_{\text{tri}}=(1)/(2)\cdot b\cdot h=(1)/(2)\cdot7\cdot8=28\; \; m^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/2d3i3f4obiamt2zf0ngkzecpizp3fi3wks.png)
So, the area of the triangle is 28 m²
The area of the shaded region can be found by subtracting the area of the triangle from the area of the rectangle.
![\begin{gathered} A=A_{\text{rect}}-A_{\text{tri}} \\ A=112-28 \\ A=84\; \; m^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/7yodgz0nejytwetga0m7g3ddzrd2nu6i30.png)
Therefore, the area of the shaded region is 84 m²