help? find the area of the regular polygon round to the nearest tenth.

Answer:
[tex]A=779.4\ cm^{2}[/tex]
Step-by-step explanation:
we know that
The area of a regular hexagon is equal to the area of six equilateral triangles
Applying the law of sines
The area of six equilateral triangles is equal
[tex]A=6[\frac{1}{2}b^{2}sin(60)][/tex]
where
b is the side length of the regular hexagon
we have
[tex]b=10\sqrt{3}\ cm[/tex]
[tex]sin(60\°)=\sqrt{3}/2\ cm[/tex]
substitute
[tex]A=6[\frac{1}{2}(10\sqrt{3})^{2}(\sqrt{3}/2)][/tex]
[tex]A=450\sqrt{3}=779.4\ cm^{2}[/tex]