Look at the picture.
The longer diagonals of the hexagon divide it into 6 equilateral triangles.
Method 1.
Use the Pythagorean theorem to calculate the height of triangle:
![a=2;\ (a)/(2)=(2)/(2)=1;\ h=?\\\\h^2+\left((a)/(2)\right)^2=a^2\\\\h^2+1^2=2^2\\h^2+1=4\ \ \ |-1\\h^2=3\to h=\sqrt3](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gjhavm36s1ar5n6kkih2dsmw2op0y5jp47.png)
Calculate the area of the triangle:
![A_\triangle=(1)/(2)\cdot2\cdot\sqrt3=\sqrt3\ ft^2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/8zy4ygfegr45i4cl16kuopksmfl1v3k64k.png)
Calculate the area of the hexagon:
![A=6A_\triangle\to A=6\cdot\sqrt3=6\sqrt3\ ft^2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/zvkf7qrc03qah6y8q7r7zjajdsln66huj7.png)
Method 2:
Use the formula of the area of an equilateral triangle:
![A_\triangle=(a^2\sqrt3)/(4)\to A_\triangle=(2^2\sqrt3)/(4)=\sqrt3\ ft^2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/z70kjvcgl9je436codkl3wfirb8aw01g60.png)
Calculate the area of the hexagon:
![A=6A_\triangle\to A=6\cdot\sqrt3=6\sqrt3\ ft^2](https://img.qammunity.org/2019/formulas/mathematics/middle-school/zvkf7qrc03qah6y8q7r7zjajdsln66huj7.png)
Method 3.
Use the trigonometric function to calculate the height of a triangle:
![\sin60^o=(h)/(a)\\\\\sin60^o=(\sqrt3)/(2)\\\\(h)/(2)=(\sqrt3)/(2)\ \ \ |\cdot2\\\\h=\sqrt3\\\\\vdots](https://img.qammunity.org/2019/formulas/mathematics/middle-school/hktr04xhdlr0eu7wrok6xlgf6gd3dcfcpt.png)