Question

A regular hexagon with a perimeter of 48 yd. Find the wares of the polygon and round to the nearest tenth.

Answer 1

Answer:

166.3 square yards

Explanation:

A regular hexagon has 6 equal sides.

Given that the perimeter = 48 yards

The side length, s:

$\begin{gathered} s=\frac{48}{6} \\ s=8\text{ yards} \end{gathered}$

Next, we use the formula below for the area of a regular hexagon:

$A=\frac{3\sqrt[]{3}}{2}s^2$

Substitute 8 for s:

$\begin{gathered} A=\frac{3\sqrt[]{3}}{2}\times8^2 \\ =96\sqrt[]{3} \\ \approx166.3\; yd^2\text{ (to the nearest tenth)} \end{gathered}$

The area of the regular hexagon is 166.3 square yards.

Method 2

A regular hexagon is divided into 6 equilateral triangles.

In this case:

The side length of the equilateral triangle = 8 yards

First, find the value of the height, h using the Pythagoras Theorem.

$\begin{gathered} 8^2=4^2+h^2 \\ h^2=8^2-4^2 \\ h^2=64-16 \\ h^2=48 \\ h=\sqrt{48} \\ h=6.93\text{ yds} \end{gathered}$

Next, find the area of one equilateral triangle:

$\begin{gathered} \text{Area}=\frac{1}{2}\times\textcolor{red}{base}\times\textcolor{green}{height} \\ =\frac{1}{2}\times\textcolor{red}{8}\times\textcolor{green}{6.93} \\ =27.72\; yd^2 \end{gathered}$

Since there are 6 equilateral triangles in the hexagon:

$\begin{gathered} \text{Area of the hexagon=6}\times Area\text{ of one equilateral triangle} \\ =6\times27.72 \\ =166.3\; yd^2 \end{gathered}$

The area of the regular hexagon is 166.3 square yards.