Question

ide of the regular hexagon.The area of a given hexagon is equal to the area of an equilateral triangle whose perimeter is 36 inches. Find the length of a side of the regular hexagon.The area of a given hexagon is equal to the area of an equilateral triangle whose perimeter is 36 inches. Find the length of a side of the regular hexagon. Please answer in square root form.

Answer 1

$\bf \textit{area of an equilateral triangle}\\\\ A=\cfrac{s^2\sqrt{3}}{4}\qquad \begin{cases} s=\textit{length of a side}\\ ----------\\ perimeter=s+s+s\\ perimiter=3s\\ 36=3s\\ \frac{36}{3}=s\\ 12=s \end{cases} \\\\\\ A=\cfrac{12^2\sqrt{3}}{4}\implies A=36\sqrt{3}$

ok.. .based on a side of 12, that's the area of the equilateral triangle, now, the hexagon has the same area... so... let's use the area of a polygon to see what's the length of a side

$\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\ cot\left( \frac{180}{n} \right)\qquad \begin{cases} n=\textit{number of sides}\\ s=\textit{length of one side}\\ ----------\\ n=6\\ A=36\sqrt{3} \end{cases} \\\\\\ 36\sqrt{3}=\cfrac{1}{4}\cdot 6\cdot s^2\cdot cot\left( \frac{180}{6} \right)\implies 36\sqrt{3}=\cfrac{1}{4}\cdot 6\cdot s^2\cdot \sqrt{3} \\\\\\ 36\sqrt{3}=\cfrac{6s^2\sqrt{3}}{4}\implies \cfrac{4\cdot 36\sqrt{3}}{6\sqrt{3}}=s^2 \\\\\\ 24=s^2\implies \sqrt{24}=s\implies 2\sqrt{6}=s$


now, in case you want to check how much is the cot(30°), check your Unit Circle, recall, cotangent is cosine/sine

Answer 2

Answer:

2squrt6 for short