Recall the formula to find the area of a regular polygon. Also, notice that the figure can be divided into equilateral triangles.
Area: 162sqrt(3) cm^2 Methods: See solution.
Practice makes perfect
Let's consider a regular hexagon of side length 6sqrt(3) cm.
We know the area of a regular polygon is given by half the product between the apothem and the perimeter.
A = 1/2* a* P
On the other hand, a regular hexagon can be divided into 6 triangles. Then, we can also find the area of the hexagon by finding the area of each triangle and multiplying it by 6.
A = 6* Area(△ ABG)
Let's perform the computations separately.
Finding the Apothem and Perimeter
We know each angle of a regular hexagon has a measure of 120^(∘), and since the radii are angle bisectors, we conclude that m∠ GAB = 60^(∘). Let's also draw the apothem, which bisects AB.
Notice that △ AGH is a 30^(∘)-60^(∘)-90^(∘) triangle. Therefore, its longer leg equals sqrt(3) times the shorter leg.
a = sqrt(3) * 3sqrt(3) ⇒
a = 9 cm
The perimeter of the regular hexagon is 6 times the length of each side.
P = 6 * 6sqrt(3) ⇒
P = 36sqrt(3) cm
Finally, we can find the area of the regular hexagon.
A = 1/2* 9 * 36sqrt(3) ⇒
A = 162sqrt(3) cm^2
Finding the Area of △ ABG
Let's draw the height of △ ABG which bisects the base AB.
As before, we have that △ AGH is a 30^(∘)-60^(∘)-90^(∘) triangle, and consequently, the longer leg is equal to sqrt(3) times the shorter leg.
GH = sqrt(3) * 3sqrt(3) ⇒
GH = 9 cm
Notice that GH is the height of △ ABG and AB is the base. Using this, let's compute the area of △ ABG.
A(△ ABG) = 6sqrt(3) * 9/2 = 27sqrt(3) cm^2
Finally, the area of the regular hexagon is equal to 6 times the area of △ ABG.
A = 6 * 27sqrt(3) ⇒
A = 162sqrt(3) cm^2
