Find the area of a regular hexagon with side length 10 m. Round to the nearest tenth.

Answer:
[tex]A=259.8\ m^2[/tex]
Step-by-step explanation:
we know that
The area of a regular hexagon is the same that the area of 6 equilateral triangles
The area of 6 equilateral triangles applying the law of sines is equal to
[tex]A=6[\frac{1}{2}b^2sin(60^o)][/tex]
where
b is the length side of the regular hexagon
we have
[tex]b=10\ m[/tex]
substitute
[tex]A=6[\frac{1}{2}(10)^2sin(60^o)][/tex]
[tex]A=259.8\ m^2[/tex]