Squares and triangles to hexagon

Relevant wiki: General Polygons - Area

By cosine law, the angles of the $\triangle GHI$ can be computed.

$(\sqrt{26})^2=(\sqrt{18})^2+(\sqrt{20})^2-2(\sqrt{18})(\sqrt{20})(\cos~\beta)$ $\implies$ $\beta=71.57$

$(\sqrt{20})^2=(\sqrt{18})^2+(\sqrt{26})^2-2(\sqrt{18})(\sqrt{26})(\cos~\theta)$ $\implies$ $\theta=56.31$

$(\sqrt{18})^2=(\sqrt{26})^2+(\sqrt{20})^2-2(\sqrt{26})(\sqrt{20})(\cos~\alpha)$ $\implies$ $\alpha=52.13$

$\angle AGF=180-56.31=123.69$ $;$ $\angle BHC=180-52.13=127.87$

$\angle EID=180-71.57=108.43$

In computing the area of each triangle, we use the formula: $\boxed{A=\dfrac{ab \sin~C}{2}}$

$A_{GHI}=\dfrac{1}{2}(\sqrt{26})(\sqrt{18})(\sin~56.31)=9$

$A_{AGF}=\dfrac{1}{2}(\sqrt{26})(\sqrt{18})(\sin~123.69)=9$

$A_{EID}=\dfrac{1}{2}(\sqrt{18})(\sqrt{20})(\sin~108.43)=9$

$A_{BHC}=\dfrac{1}{2}(\sqrt{26})(\sqrt{20})(\sin~127.87)=9$

$A_{ABCDEF}=26+18+20+4(9)=$ $\boxed{100}$

Rotate the inner triangle by 60 degrees about any of its vertices, coming into contact with an outer triangle. Then since both triangles share the same altitude and base length, they have the same area. Using Heron's formula determines the area of the inner triangle to be 9. Hence, the total area is 4(9) + 18 + 20 + 26 = 100.