Calculate the area of the following regular octagon.

According to the given figure in question,

Area of the regular octagon = 

Area of trapezium ABCD + area of rectangle AHED + area of trapezium HGFE  

 $=\frac{1}{2}\times (\mathrm{AD}+\mathrm{BC})\times \mathrm{h}+(\mathrm{AD}\times \mathrm{DE})+\frac{1}{2}\times \left(\mathrm{GF}+\mathrm{EH}\right)\times \mathrm{h}$

AB = BC = CD = DE = EF = FG = GH = AH = 5 m as it is a regular octagon all sides are equal

from figure AD = EH = 11 m and height of the trapezium h = 4 m

$$\begin{gathered} = \frac{1}{2}\times \left(11+5\right)\times 4+\left(11\times 5\right)+\frac{1}{2}\times \left(5+11\right)\times 4 \\ =32+55+32 \\ =119 {\mathrm{m}}^2 \end{gathered}$$

Therefore, the area of the regular octagon ABCDEFGH = 119 m²