Third or Sixth Root Unity?

Area of $\triangle ABG = \dfrac {1}{2} \times AB \times AE = \dfrac {1}{2} \times 2 \times 2 \times AF \times \cos{30^\circ}$

$\quad \quad \quad \quad \quad \quad \space = \dfrac {1}{2} \times 2 \times 2 \times 2 \times \dfrac {\sqrt{3}}{2} = 2\sqrt{3} = \boxed {3.464}$

Look at the figure below: In the figure is the hexagon $ABCDEF$ and in the second figure it is also divided into 6 equilateral triangles.The height of the equilateral triangles divides them into 2 right triangles with base 1 cm and hypotenuse 2 cm.Therefore by the Pythagorean theorem we get: $(Height)^2+1^2=2^2\\(Height)^2=4-1=3\\Height=\sqrt{3}$

The height of the equilateral triangles is half the height of $ABG$ .So the height of $ABG$ is $2\times\sqrt{3}=2\sqrt{3}$ .The base of $ABG$ is 2cm so: $Area=\frac{1}{2}\times breadth \times height\\=\frac{1}{2}\times2\times2\sqrt{3}\\=2\sqrt{3}=3.464\;To\;4\;decimal\;places$

In $\bigtriangleup BCD -$

By sine rule,

$\frac{BD}{\sin120} = \frac{BC}{\sin30}$

$BD = 4 \times \sin120$

Area of Shaded $\bigtriangleup = \frac{1}{2} \times DE \times BD$

$= \frac{1}{2} \times 2 \times 4 \times \sin120$

$= 4 \times 0.866$

$= 3.464$

$CD=AE=1,AB=2AD=2\sqrt{AC^2-AD^2}=2\sqrt{3}$ $∆AEF=\dfrac{1}{2}\times AB\times BF=\dfrac{1}{2}\times 2\sqrt{3}\times 2=2\sqrt{3}=\boxed{3.464}$

The apothem is calculated with the formula sqrt(side^2 - (side/2)^2)

sqrt(2^2 - (2/2)^2) = sqrt(3)

The height of the triangule is 2 * apothem => 2 sqrt(3)

The area of a triangle is calculated with the formula (base * height)/2

=> 2*2sqrt(3)/2 = 2 sqrt(3)

Ans: 2 sqrt(3)

Area of ABG = 1/2(AB)(h) join E and A and h=EA such that angle F meets EA at O, F is divided in two angles F= EFO+AFO=120 (Hexagon's angle) EFO=30 and AFO=90 EO=FE Cos30 , OA= FA Cos30 EO=2(0.866), OA=2(0.866) EO=1.732, OA=1.732 EA=EO+OA EA=3.464 A=1/2(h)(AB) A=1/2(3.464)(2) A=3.464 sq cm

A=1/2 bh; a=1cot 30=sqrt3 and h=2a Substituting, A=1/2 x 2 x 2sqrt3 =2x sqrt3=3.464sqcm.