Find the area of the shaded field shown in the figure.
(N/A) In the given figure,join $DE$.
From the figure,the diameter of the semi-circle $DFE$ is $6 - 4 = 2 \, m$. Therefore,the radius $r = 2 / 2 = 1 \, m$.
Now,the area of the rectangle $ABCD$ is $BC \times AB = 8 \times 4 = 32 \, m^2$.
The area of the semi-circle $DFE$ is $\frac{\pi r^2}{2} = \frac{\pi}{2}(1)^2 = 0.5 \pi \, m^2$.
Therefore,the total area of the shaded region = Area of rectangle $ABCD$ + Area of semi-circle $DFE$.
Total area = $(32 + 0.5 \pi) \, m^2$.
From the figure,the diameter of the semi-circle $DFE$ is $6 - 4 = 2 \, m$. Therefore,the radius $r = 2 / 2 = 1 \, m$.
Now,the area of the rectangle $ABCD$ is $BC \times AB = 8 \times 4 = 32 \, m^2$.
The area of the semi-circle $DFE$ is $\frac{\pi r^2}{2} = \frac{\pi}{2}(1)^2 = 0.5 \pi \, m^2$.
Therefore,the total area of the shaded region = Area of rectangle $ABCD$ + Area of semi-circle $DFE$.
Total area = $(32 + 0.5 \pi) \, m^2$.
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