Answer:
[tex]A=259.8\ cm^2[/tex]
Step-by-step explanation:
we know that
The figure represent a regular hexagon
A regular hexagon can be divided into six congruent equilateral triangles
Let
b ---> the length side of the regular hexagon
see the attached figure to better understand the problem
[tex]cos(30\°)=\frac{5\sqrt{3}}{b}[/tex]
Remember that
[tex]cos(30\°)=\frac{\sqrt{3}}{2}[/tex]
so
[tex]\frac{5\sqrt{3}}{b}=\frac{\sqrt{3}}{2}\\\\b=10\ cm[/tex]
Find the area of the regular hexagon
The area of a regular hexagon is equal to the area of six congruent equilateral triangles
[tex]A=6[\frac{1}{2}(b)(h)][/tex]
we have
[tex]b=10\ cm[/tex]
[tex]h=5\sqrt{3}\ cm[/tex]
substitute
[tex]A=6[\frac{1}{2}(10)(5\sqrt{3})][/tex]
[tex]A=150\sqrt{3}\ cm^2[/tex]
[tex]A=259.8\ cm^2[/tex]
