Calculating the area of a hexagon (6-sided entity) alternative

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I am kinda curious in something. When looking at a hexagon (6-sided entity) and wanting to calculate the area, could you in theory treat the hexagon as 4 triangles and one square in the middle like this?

Here is my calculations, if this can be done
$\left(\frac{C×A/2}{1/2}×4\right) + (B×A)$

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