Area of a regular hexagon

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So i am trying to brush up on some basic polygon mathematics and i have this question that i am having trouble visualizing. What would be the area of a hexagon of side length 4. Any ideas on how to do this?

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Hint: Divide it into $6$ congruent triangles. To get the interior angle of the hexagon, use the formula $\frac{180}n(n-2)$. Inserting $6$, we get, $30*4=120^{\circ}$. And the two triangles are congruent, which shows that all the base angles in all the triangles are $60^{\circ}$, thus, all are equilateral. Now we have the side length $4$, now by using the formula, $\frac{\sqrt3a^2}{4}$, we get the area, $$\boxed{6.\frac{\sqrt3.16}{4}=24\sqrt3}$$