Join the center of the hexagon with the 2 base angles.
An equilateral triangle, with side length x, is formed.
(remark: a regular hexagon is made up of 6 equilateral triangles with equal length)
The height [tex]2 \sqrt{3} [/tex] forms 2 congruent right triangles with :
hypotenuse= x, side_1=x/2, and side_2= [tex]2 \sqrt{3} [/tex].
From the pythagorean theorem we have:
[tex] x^{2} = ( \frac{x}{2} )^{2}+(2 \sqrt{3})^{2} [/tex]
[tex]x^{2} = \frac{ x^{2} }{4} +12[/tex]
[tex] \frac{3}{4} x^{2} =12[/tex]
[tex] x^{2} = \frac{12*4}{3}=4*4 [/tex]
thus, x=4.
The area of the triangle is 1/2 * 4 * [tex]2 \sqrt{3}[/tex]=6.93 (mm squared)
The area of the hexagon is 6* the area of the triangle = 42 (mm squared)
Answer: a. 42 (mm squared)
