Question

A regular hexagon has sides of 2 feet. What is the area of the hexagon?

Answer 1

Trigonometry would help with this question.

The area of a regular hexagon is ((3√3)s^2)/2 where s is the side.

Plugging in 2 gives us 6√3 or 10.39 feet.

Answer 2

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:

$a=2;\ \dfrac{a}{2}=\dfrac{2}{2}=1;\ h=?\\\\h^2+\left(\dfrac{a}{2}\right)^2=a^2\\\\h^2+1^2=2^2\\h^2+1=4\ \ \ |-1\\h^2=3\to h=\sqrt3$

Calculate the area of the triangle:

$A_\triangle=\dfrac{1}{2}\cdot2\cdot\sqrt3=\sqrt3\ ft^2$

Calculate the area of the hexagon:

$A=6A_\triangle\to A=6\cdot\sqrt3=6\sqrt3\ ft^2$

Method 2:

Use the formula of the area of an equilateral triangle:

$A_\triangle=\dfrac{a^2\sqrt3}{4}\to A_\triangle=\dfrac{2^2\sqrt3}{4}=\sqrt3\ ft^2$

Calculate the area of the hexagon:

$A=6A_\triangle\to A=6\cdot\sqrt3=6\sqrt3\ ft^2$

Method 3.

Use the trigonometric function to calculate the height of a triangle:

$\sin60^o=\dfrac{h}{a}\\\\\sin60^o=\dfrac{\sqrt3}{2}\\\\\dfrac{h}{2}=\dfrac{\sqrt3}{2}\ \ \ |\cdot2\\\\h=\sqrt3\\\\\vdots$